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01Sequences: convergence, limit laws, and the monotonic sequence theorem

A patient takes one $150$ mg tablet every morning, and by the next morning the body has cleared exactly half of whatever drug was in the blood. Track the amount right after each dose: $150$, then $225$, then $262.5$, then $281.25\ldots$ Every number is larger than the last, so does the drug pile up without limit, or does it level off at some safe ceiling?

By the end of this section you can prove the amounts stay under $300$ mg forever, prove they keep rising, and conclude from those two facts alone that they close in on exactly $300$ mg.

In 60 seconds

A sequence is an unending list $a_1,a_2,a_3,\ldots$; it converges to $L$ when its terms eventually stay as close to $L$ as you like, and the fastest way to decide is to reuse everything you know about limits of functions.

Limit from a function
$$\lim_{x\to\infty}f(x)=L,\ a_n=f(n)\ \Rightarrow\ \lim_{n\to\infty}a_n=L$$

any $a_n$ you can read off a function of a real $x$; unlocks L'Hopital

$$b_n\le a_n\le c_n,\ \lim b_n=\lim c_n=L\ \Rightarrow\ \lim a_n=L$$

a bounded wobble such as $\cos n$ or $(-1)^n$ divided by something growing

Limit of a geometric sequence
$$\lim_{n\to\infty}r^{\,n}=0\ (|r|<1),\quad =1\ (r=1),\quad \text{diverges otherwise}$$

anything of the form $r^n$, or ratios of powers like $2^n/5^n$

Monotonic Sequence Theorem
$$\text{bounded}+\text{monotonic}\ \Rightarrow\ \text{convergent}$$

recursively defined $a_{n+1}=f(a_n)$ where no formula for $a_n$ is available

Three most common mistakes

  1. Reading off a limit from the first few terms. Finitely many terms never decide a limit; $a_n=\cos(1/n)$ starts near $1$ and stays there, but a lookalike could peel away at $n=10^6$.

  2. Writing $\lim(-1)^n=1$ because the magnitude is $1$. The size settling down is not the sequence settling down; the even and odd terms head to $+1$ and $-1$, so there is no limit.

  3. Solving $L=f(L)$ and calling it the limit without first proving the sequence converges. A divergent recursion also satisfies that equation, so the fixed point is only the answer once convergence is established.

Only the quiz weight (10%) and homework weight (10%) are published for this course; sequence limits are exactly the short, self-checking computations those quizzes and homework sets are built from, so treat every worked example here as a graded item.

How much time do you have?
10 minutes

The first line of almost every question: divide by the highest power of $n$, or recognise an $r^n$, and read the limit straight off.

card, c-limit-laws, c-geometric-and-key-limits, formula card
45 minutes

Every type that can be set: the convergence test, the four computational tools, the squeeze for wobbling terms, and the two-step monotonic argument for recursions.

card, c-limit-of-sequence, c-limit-laws, c-squeeze-and-abs, c-geometric-and-key-limits, c-monotonic-bounded, faded ladder, formula card
full read

Where sequences come from, why each tool is legal, and enough mixed practice that you pick the method yourself instead of being handed it.

hook, prerequisites, c-sequence-basics, c-limit-of-sequence, c-limit-laws, c-squeeze-and-abs, c-geometric-and-key-limits, c-monotonic-bounded, contrast pairs, faded ladder, exam example, practice A to D
By the end of this section

  1. Represent a sequence by an explicit formula, by a recursion, and as points plotted at the integers, and read terms off each form.

  2. Decide whether a sequence converges and state its limit, using the meaning of convergence rather than a table of terms.

  3. Compute limits of sequences with the limit laws and by passing to a function of a real variable so that L'Hopital's Rule applies.

  4. Apply the Squeeze Theorem, the absolute-value test for a zero limit, and the continuous-function theorem to sequences with a bounded oscillating part.

  5. Determine the limit of a geometric sequence and of the standard sequences, and use them inside larger limits.

  6. Prove that a bounded monotonic sequence converges, and find the limit of a once convergence is secured.

Syllabus coverage
11.1

Sequences

The whole of this section is section 11.1: what a sequence is, when it converges, the tools that compute the limit, and the monotonic sequence theorem for recursions. Series themselves begin in the next section.

covered
Recall first
Limit at infinity of a rational function

Divide numerator and denominator by the highest power of $x$; every $1/x^k$ then tends to $0$, leaving the ratio of leading coefficients.

Exactly this move computes the limit of any rational sequence, with $n$ in the role of $x$.

L'Hopital's Rule

If $\lim f/g$ is of the form $0/0$ or $\infty/\infty$, then $\lim \dfrac{f}{g}=\lim \dfrac{f'}{g'}$ provided the right-hand limit exists.

Sequences like $(\ln n)/n$ have no algebraic simplification; the only route is to differentiate a matching function of $x$.

Limit Laws

For limits that exist, the limit of a sum, difference, product, quotient (nonzero denominator) and power is obtained termwise.

The same laws hold verbatim for sequences and justify breaking a limit into manageable pieces.

The number $e$ as a limit

$\displaystyle\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{x}=e^{a}$.

A recurring sequence limit is exactly this form with $x=n$, and no other tool touches it cleanly.

Try it yourself first (3 questions)
1§01.2 — approaching a value versus reaching it●●○○○

A classmate revising for the quiz insists that if a sequence has a limit $L$, then one of its terms has to actually equal $L$ at some point, otherwise there is nothing for the terms to reach. It sounds reasonable, and it is worth settling before we build anything on the word limit.

Given

  • Claim: if $\lim_{n\to\infty}a_n=L$, then $a_n=L$ for at least one $n$.

Find

  1. (a) Is the claim true or false? Give a sequence that decides it.

such as $1/n$ can approach its limit $0$ without any single term ever being equal to $0$.

Hint 1/4

Separate two ideas that the claim glues together: getting arbitrarily close to $L$, and equalling $L$.

Hint 2/4

Convergence asks that the terms eventually stay within any band around $L$; it never asks any term to land on $L$.

Hint 3/4

Test it on $a_n=1/n$: the terms are $1,\tfrac12,\tfrac13,\ldots$ closing in on $0$, yet no term is ever $0$.

Hint 4/4

So the claim is false: $1/n\to 0$ while every term is strictly positive.

Show solution
Produce a counterexample
$$a_n=\tfrac1n,\quad \lim_{n\to\infty}\tfrac1n=0$$

the terms shrink through every positive value toward $0$

$$a_n=\tfrac1n>0\ \text{for all }n$$

no term is ever $0$, so the limit $0$ is approached but never attained

Answer $$\text{False}$$
Check

The graph of $1/n$ never touches the axis, yet the dots drop below any positive height you name.

Limit is about the destination the terms crowd toward, not a place they must arrive at.

2§01.3 — a limit at infinity recalled from Calculus I●○○○○

Before sequences we need one reflex from Calculus I fully awake: the limit of a rational function as the variable runs to infinity. We will use the very same move on sequences, just with $n$ in place of $x$.

Given

  • $f(x)=\dfrac{2x+1}{x+3}$

Find

  1. (a) Compute $\lim_{x\to\infty}f(x)$.

Hint 1/4

The value is controlled by the largest power of $x$ on top and bottom, here $x^1$ in both.

Hint 2/4

Divide every term of numerator and denominator by that highest power, $x$, then send each $1/x$ to $0$.

Hint 3/4

With $f(x)=\dfrac{2x+1}{x+3}$: dividing by $x$ gives $\dfrac{2+1/x}{1+3/x}$, and $1/x\to 0$.

Hint 4/4

So the limit is $2/1=2$.

Show solution
Divide by the highest power
$$\frac{2x+1}{x+3}=\frac{2+\frac1x}{1+\frac3x}$$

dividing top and bottom by $x$ leaves constants plus terms that vanish

$$\lim_{x\to\infty}\frac{2+\frac1x}{1+\frac3x}=\frac{2+0}{1+0}=2$$

each $1/x$ tends to $0$, so only the leading coefficients survive

Answer $$\boxed{2}$$
Check

At $x=1000$, $f=2001/1003=1.995\ldots$, already within $0.005$ of $2$.

3§01.3 — L'Hopital on an infinity over infinity form●●○○○

One more Calculus I tool has to be sharp, because it is the only way to handle a logarithm growing against a power. If direct substitution gives $\infty/\infty$, L'Hopital's Rule lets you differentiate top and bottom separately.

Given

  • $g(x)=\dfrac{\ln x}{x}$

Find

  1. (a) Compute $\lim_{x\to\infty}g(x)$ using L'Hopital's Rule.

Hint 1/4

Check the form first: as $x\to\infty$ both $\ln x$ and $x$ run to $\infty$, so it is $\infty/\infty$.

Hint 2/4

L'Hopital's Rule for $\infty/\infty$: replace the ratio by the ratio of derivatives, $\dfrac{(\ln x)'}{(x)'}$.

Hint 3/4

Here $(\ln x)'=1/x$ and $(x)'=1$, so the new ratio is $\dfrac{1/x}{1}=\dfrac1x$, evaluated as $x\to\infty$.

Hint 4/4

Since $1/x\to 0$, the original limit is $0$.

Show solution
Confirm the form, then differentiate top and bottom
$$\lim_{x\to\infty}\frac{\ln x}{x}=\frac{\infty}{\infty}$$

an indeterminate form, so L'Hopital's Rule is licensed

$$\overset{\text{L'H}}{=}\lim_{x\to\infty}\frac{1/x}{1}=\lim_{x\to\infty}\frac1x=0$$

differentiate numerator and denominator once; the result is elementary

Answer $$\boxed{0}$$
Check

At $x=100$, $(\ln 100)/100=0.046$, and it keeps falling; the power outpaces the logarithm.

Carry this exact result into the sequence world: $(\ln n)/n\to 0$ for the same reason.

Notation
symbolreads asmeanswatch out
$\{a_n\}$

the sequence a-sub-n

the whole ordered list $a_1,a_2,a_3,\ldots$, one term for each positive integer $n$

The braces here do not mean a set: order matters and repeats are allowed, unlike $\{1,2\}$ as a set.

$a_n$

a-sub-n, the n-th term

the single number sitting in position $n$ of the list

$a_n$ is one number; $\{a_n\}$ is the entire sequence. Keep the two apart.

$\displaystyle\lim_{n\to\infty}a_n=L$

the limit of a-sub-n as n goes to infinity equals L

the terms eventually stay arbitrarily close to $L$

$n\to\infty$ only: $n$ is a positive integer, so there is no $n\to 0$ or $n\to a$ for a sequence.

$a_{n+1}=f(a_n)$

a-sub-n-plus-one equals f of a-sub-n

a recursion: each term is built from the one before by the rule $f$

You also need a starting value $a_1$; the recursion alone does not pin down the sequence.

$n!$

n

the product $1\cdot 2\cdot 3\cdots n$ of all integers up to $n$

$n!$ grows faster than any fixed power or any $r^n$; that ranking decides many limits.

Conventions used here
What we mean by diverges, and the shorthand for growth to infinity.

A sequence converges when it has a single finite limit; otherwise it diverges. Divergence covers two very different pictures: terms that grow without bound, and terms that oscillate. When $a_n$ grows past every bound we write $\lim_{n\to\infty}a_n=\infty$ as a description of how it diverges, but this sequence is still divergent; $\infty$ is not a value the sequence attains or converges to.

Students routinely tick converges for a sequence heading to $\infty$ because a limit symbol got written. Fixing the vocabulary once stops that: $\lim=\infty$ is a divergence, in words, not a convergence.

How an alternating sign is reported.

A sequence like $(-1)^n$ diverges by oscillation: its limit does not exist because the even terms sit at $+1$ and the odd terms at $-1$. We never write $\lim(-1)^n=\pm 1$; two different subsequential values is precisely the definition of no limit.

The magnitude of $(-1)^n$ is a constant $1$, which tempts the answer $1$. Naming this a divergence keeps the size of the terms and the limit of the terms as separate questions.

The boundary cases of $r^{\,n}$.

For $r^n$ the dividing lines matter: $r=1$ gives the constant sequence $1,1,1,\ldots$ which converges to $1$, while $r=-1$ gives $-1,1,-1,\ldots$ which diverges. So $r^n$ converges exactly when $-1<r\le 1$, and its limit there is $0$ for $|r|<1$ and $1$ for $r=1$.

The two endpoints behave oppositely and are the most common slip; stating them explicitly removes the guesswork at $r=\pm 1$.

Indexing and the argument of trig functions.

Unless stated otherwise a sequence starts at $n=1$, and shifting the start changes no limit. Inside a sequence, $\sin n$ and $\cos n$ take $n$ in radians, and these values do not settle down; only their being trapped in $[-1,1]$ is used.

A late start index or a stray degree reading changes numbers without changing conclusions, so we fix both up front and lean only on the bound $|\sin n|\le 1$.

1.1A sequence is a list indexed by the integers

Fixes what a sequence is (a function on the integers), how it is given, and why finitely many terms cannot decide its limit.

In Calculus I every limit ran along a continuous curve, $x\to a$ or $x\to\infty$. A sequence lives only at the whole numbers $1,2,3,\ldots$, and that one change is what the chapter turns on.

Solvable with what we have

  • List as many dosing amounts as we like: $150,\ 225,\ 262.5,\ 281.25,\ 290.625,\ldots$

  • Compute any single term on demand, even the $100$th.

  • Plot the terms as separate dots above $n=1,2,3,\ldots$

Not solvable yet

  • Say for certain what number, if any, the list is heading toward.

  • Rule out that the amounts creep past $300$ and keep going.

  • Turn from every term is bigger than the last into it must settle down.

Read the trend off the first few terms:

$$150,\ 225,\ 262.5,\ 281.25,\ 290.625\ \longrightarrow\ \text{looks like about }300.$$

The gaps are shrinking, so eyeballing suggests $300$. But eyeballing five terms cannot tell $300$ apart from $305$, and a sequence can look calm for a million terms and then move.

Why it fails

A finite list is compatible with many different destinations, so compute more terms is evidence, never proof. We need a precise meaning for the terms settle down (the next section) and a theorem that certifies it from structure rather than from a table (the last section).

DefinitionDefinition: Sequence
Conditions

  • the domain is the positive integers $1,2,3,\ldots$ (occasionally starting at $0$)

  • a term may be given by an explicit formula $a_n$, or by a recursion $a_{n+1}=f(a_n)$ with a stated first term

  • order and repetition carry information: a sequence is not the set of its values

$$\boxed{\;\{a_n\}:\ a\colon\{1,2,3,\ldots\}\to\mathbb{R},\qquad a_n=a(n)\;}$$

A sequence is nothing but a function whose inputs are the counting numbers; the output at input $n$ is the $n$-th term. Writing $\{a_n\}$ just lists those outputs in order.

Looks like this, but is not

This is a sequence: $1,2,1,2,1,2,\ldots$, that is $a_n=1$ for odd $n$ and $a_n=2$ for even $n$.

This looks like the same object but is not: the set $\{1,2\}$. The set forgets order and forgets that each value recurs forever; the sequence remembers both. That is why $\{(-1)^n\}$ can diverge even though the set of its values is just $\{-1,1\}$.

First terms of $a_n=(-1)^n\,\dfrac{n}{n+1}$

Write the first four terms of $a_n=(-1)^n\dfrac{n}{n+1}$ and describe the pattern.

Given

  • $a_n=(-1)^n\dfrac{n}{n+1}$

Find

The terms $a_1,\ldots,a_4$ and the trend.

Solution
Substitute $n=1,2,3,4$
$$a_1=(-1)^1\tfrac{1}{2}=-\tfrac12,\quad a_2=(+1)\tfrac{2}{3}=\tfrac23$$

the factor $(-1)^n$ flips sign each step; the size is $n/(n+1)$

$$a_3=-\tfrac34,\quad a_4=\tfrac45$$

the magnitudes $\tfrac12,\tfrac23,\tfrac34,\tfrac45$ climb toward $1$

Answer $$\{a_n\}=-\tfrac12,\ \tfrac23,\ -\tfrac34,\ \tfrac45,\ \ldots$$
Check

The signs read $-,+,-,+$, matching $(-1)^n$ starting at $n=1$; the sizes match $n/(n+1)$ term by term.

The sign and the size are two separate stories; later we decide the limit by handling them separately.

Unfolding the dosing recursion

The dosing model is $a_1=150$ and $a_{n+1}=\tfrac12 a_n+150$. Compute $a_2,a_3,a_4$.

Given

  • $a_1=150$

  • $a_{n+1}=\tfrac12 a_n+150$

Find

The next three terms.

Solution
Apply the rule repeatedly
$$a_2=\tfrac12(150)+150=225$$

feed $a_1$ into the rule to get $a_2$

$$a_3=\tfrac12(225)+150=262.5$$

now feed $a_2$ back in; each output becomes the next input

$$a_4=\tfrac12(262.5)+150=281.25$$

one more turn of the same crank

Answer $$a_2=225,\ a_3=262.5,\ a_4=281.25$$
Check

Each term is halfway between the previous term and $300$, so the gap to $300$ halves every step: $150,75,37.5,18.75$.

A recursion gives terms cheaply but hides the destination; that is exactly the gap the last section closes.

Checkpoint
§01.1 — reading terms off a formula●○○○○

Stay thirty seconds on a sequence that is neither purely rising nor purely falling. It is $a_n=3+\dfrac{(-1)^n}{n}$, a steady $3$ with a shrinking wobble on top.

Given

  • $a_n=3+\dfrac{(-1)^n}{n}$

Find

  1. (a) Write $a_1,a_2,a_3$ and say whether each sits above or below $3$.

Hint 1/4

The $3$ is fixed; only the $\dfrac{(-1)^n}{n}$ piece moves the term off $3$.

Hint 2/4

For odd $n$ that piece is negative (below $3$); for even $n$ it is positive (above $3$).

Hint 3/4

Substitute: $a_1=3-1$, $a_2=3+\tfrac12$, $a_3=3-\tfrac13$.

Hint 4/4

So $a_1=2$ (below), $a_2=3.5$ (above), $a_3=\tfrac83\approx2.667$ (below).

Show solution
Evaluate and compare with 3
$$a_1=3+\tfrac{-1}{1}=2$$

odd $n$ makes the correction negative, so below $3$

$$a_2=3+\tfrac{1}{2}=3.5$$

even $n$ makes it positive, so above $3$

$$a_3=3+\tfrac{-1}{3}=\tfrac83\approx2.667$$

odd again, below $3$, but closer than $a_1$ was

Answer $$\boxed{a_1=2,\ a_2=3.5,\ a_3\approx2.667}$$
Check

The distances from $3$ are $1,\tfrac12,\tfrac13$, shrinking, which foreshadows a limit of $3$.

⚠ Treating a sequence as the set of its values

the brace notation $\{a_n\}$ looks exactly like set notation learned earlier

wrong$$\{(-1)^n\}=\{-1,1\}\ \Rightarrow\ \text{two values, so it converges to one of them}$$
right$$\{(-1)^n\}=-1,1,-1,1,\ldots\ \text{(order kept, so it can diverge)}$$
⚠ Thinking a recursion alone defines the sequence

the rule $a_{n+1}=f(a_n)$ looks complete on its own

wrong$$a_{n+1}=\tfrac12 a_n+150\ \Rightarrow\ \text{one sequence}$$
right$$a_{n+1}=\tfrac12 a_n+150,\ a_1=150\ \Rightarrow\ \text{now pinned down}$$

1.2Convergence: the terms eventually stay inside any band

Gives the exact meaning of converges: past some point every term sits inside any band you name around the limit.

We can list terms and plot them. Now we pin down, with no hand-waving, what it means for the list to have a destination $L$.

DefinitionDefinition: (convergence)
Conditions

  • $L$ is a single finite number

  • the challenge $\varepsilon>0$ is arbitrary, however small

  • the response $N$ may depend on $\varepsilon$; only the tail $n>N$ must obey

$$\boxed{\;\lim_{n\to\infty}a_n=L\iff \forall\varepsilon>0\ \exists N\ \text{s.t.}\ n>N\Rightarrow |a_n-L|<\varepsilon\;}$$

Name any band of half-width $\varepsilon$ around $L$. The sequence converges to $L$ if, no matter how narrow that band, all terms from some point $N$ onward fall inside it. If no single $L$ works, the sequence diverges.

Looks like this, but is not

This looks like it should converge: $(-1)^n$ is bounded, never escaping $[-1,1]$, and its terms have constant size $1$.

But bounded is not the same as convergent: try any candidate $L$. A band of half-width $\tfrac12$ around $L$ cannot contain both $+1$ and $-1$ (they are $2$ apart), so from no point on do all terms sit inside it. No $L$ survives, so $(-1)^n$ diverges.

Proving $\dfrac{n}{n+1}\to 1$ from the definition

Show $\dfrac{n}{n+1}\to 1$, and for the band $\varepsilon=0.01$ give an $N$ that works.

Given

  • $a_n=\dfrac{n}{n+1}$

  • $L=1$

  • $\varepsilon=0.01$

Find

A proof, plus an explicit $N$ for $\varepsilon=0.01$.

Solution
Measure the distance to the target
$$|a_n-1|=\left|\tfrac{n}{n+1}-1\right|=\left|\tfrac{-1}{n+1}\right|=\tfrac{1}{n+1}$$

combine over a common denominator; the gap is a clean $1/(n+1)$

Force the distance below $\varepsilon$
$$\tfrac{1}{n+1}<\varepsilon\iff n+1>\tfrac1\varepsilon\iff n>\tfrac1\varepsilon-1$$

solve the band condition for $n$; this names the entry point

$$\varepsilon=0.01:\ n>100-1=99,\ \text{take }N=99$$

any $n>99$ then satisfies $|a_n-1|<0.01$

Answer $$\boxed{\lim_{n\to\infty}\tfrac{n}{n+1}=1,\quad N=99\ \text{for}\ \varepsilon=0.01}$$
Check

Check the boundary: $a_{100}=100/101=0.990099$, and $|0.990099-1|=0.0099<0.01$, while $a_{99}=0.99$ sits exactly on the edge.

The pattern $N\approx 1/\varepsilon$ is typical: halving the band roughly doubles how long you wait.

Proving $(-1)^n$ diverges

Show that $(-1)^n$ has no limit.

Given

  • $a_n=(-1)^n$

Find

A proof that no $L$ works.

Solution
Assume a limit and derive a contradiction
$$\text{Suppose }\lim(-1)^n=L,\ \text{take }\varepsilon=\tfrac12$$

a genuine limit must survive every band, so we are free to pick a small one

$$|1-L|<\tfrac12\ \text{and}\ |-1-L|<\tfrac12$$

the tail contains both even terms ($+1$) and odd terms ($-1$)

$$2=|1-(-1)|\le|1-L|+|L-(-1)|<\tfrac12+\tfrac12=1$$

triangle inequality; but $2<1$ is false

Answer $$\boxed{(-1)^n\ \text{diverges}}$$
Check

The contradiction $2<1$ came from assuming a single $L$; two clustering values is exactly the failure of a limit.

Notice this never used a formula for $L$: the argument rules out all $L$ at once.

Checkpoint
§01.2 — how many terms miss the band●●○○○

Back to $n/(n+1)\to 1$, but widen the band to $\varepsilon=0.1$. The point of the definition is that only a finite head of the sequence ever misses the band.

Given

  • $a_n=\dfrac{n}{n+1}\to 1$

  • band half-width $\varepsilon=0.1$

Find

  1. (a) How many terms $a_n$ satisfy $|a_n-1|\ge 0.1$?

Hint 1/4

The terms outside the band are exactly those whose distance to $1$ is at least $0.1$.

Hint 2/4

That distance is $|a_n-1|=\dfrac{1}{n+1}$, so you need $\dfrac{1}{n+1}\ge 0.1$.

Hint 3/4

Solve $\dfrac{1}{n+1}\ge 0.1$: it gives $n+1\le 10$, that is $n\le 9$.

Hint 4/4

So $n=1,2,\ldots,9$ miss the band: exactly $9$ terms.

Show solution
Turn the band condition into an inequality in $n$
$$\tfrac{1}{n+1}\ge 0.1\iff n+1\le 10\iff n\le 9$$

reciprocate the inequality; the direction flips because both sides are positive

$$n\in\{1,2,\ldots,9\}$$

these are the integers that violate the band

Answer $$\boxed{9\ \text{terms}}$$
Check

The tenth term $a_{10}=10/11=0.909$ has distance $0.0909<0.1$, so it is the first one inside.

⚠ Concluding a limit from the first several terms

the early terms often march so steadily that the destination looks obvious

wrong$$0.9,0.99,0.999,\ldots\ \Rightarrow\ \text{limit is }1\ \text{(by inspection)}$$
right$$\text{prove the tail enters every band: }|a_n-1|<\varepsilon\ \text{for }n>N$$
⚠ Assuming a must converge

bounded feels like tamed, and tame sequences seem to settle

wrong$$(-1)^n\in[-1,1]\ \Rightarrow\ \text{converges}$$
right$$(-1)^n\ \text{is bounded but diverges; boundedness alone is not enough}$$

1.3Computing limits: the laws and the jump to a function

Turns the definition into fast computation: reuse the limit laws, and jump to a function of $x$ so L'Hopital's Rule applies.

Chasing $N$ by hand proves a limit but is far too slow to compute one. The limit laws carry over unchanged, and one theorem lets Calculus I do the heavy lifting.

TheoremTheorem: a sequence read off a function
Conditions

  • there is a function $f$ with $a_n=f(n)$ for all $n$

  • $\lim_{x\to\infty}f(x)$ exists (a finite $L$)

  • the arrow runs one way only: from the function to the sequence, never back

$$\boxed{\;\lim_{x\to\infty}f(x)=L\ \text{and}\ a_n=f(n)\ \Longrightarrow\ \lim_{n\to\infty}a_n=L\;}$$

If the continuous graph settles on $L$ as $x$ runs to infinity, then the dots at the integers, being points on that graph, settle on $L$ too. This is what licenses L'Hopital's Rule on sequences, once you rewrite $n$ as a real $x$.

Looks like this, but is not

This looks like the theorem should run in reverse: $a_n=\sin(\pi n)=0$ for every integer $n$, so $a_n\to 0$; surely then $f(x)=\sin(\pi x)\to 0$?

The reverse direction is false: $f(x)=\sin(\pi x)$ oscillates forever and has no limit at infinity, even though it happens to be $0$ at every integer. A sequence can converge while its parent function diverges, so you may pass from $f$ to $a_n$ but not from $a_n$ back to $f$.

na_ndistance to 3

10

3.232

0.232

100

3.020

0.020

1000

3.0020

0.0020

10000

3.00020

0.00020

Each tenfold increase in $n$ shaves the distance to $3$ by roughly a factor of ten, which is the signature of a $1/n$-sized error.

Limit of $\dfrac{3n^2+2n}{n^2-1}$

Compute $\displaystyle\lim_{n\to\infty}\frac{3n^2+2n}{n^2-1}$.

Given

  • $a_n=\dfrac{3n^2+2n}{n^2-1}$

Find

The limit.

Solution
Divide by the highest power of $n$
$$\frac{3n^2+2n}{n^2-1}=\frac{3+\frac{2}{n}}{1-\frac{1}{n^2}}$$

dividing top and bottom by $n^2$ turns growing terms into vanishing ones

Send the small terms to zero
$$\lim_{n\to\infty}\frac{3+\frac2n}{1-\frac1{n^2}}=\frac{3+0}{1-0}=3$$

$2/n\to0$ and $1/n^2\to0$ by the limit laws; only leading coefficients remain

Answer $$\boxed{3}$$
Check

The table confirms it: at $n=1000$ the value is $3.0020$, matching a $1/n$-sized gap above $3$.

For any ratio of polynomials of equal degree, the limit is just the ratio of the leading coefficients.

Limit of $\dfrac{\ln n}{n}$ via a function

Compute $\displaystyle\lim_{n\to\infty}\frac{\ln n}{n}$.

Given

  • $a_n=\dfrac{\ln n}{n}$

Find

The limit.

Solution
Pass to a real variable so calculus applies
$$a_n=f(n),\quad f(x)=\frac{\ln x}{x}$$

there is no algebra to simplify $\ln n$, so we work with the function $f$ instead

Use L'Hopital on the $\infty/\infty$ form
$$\lim_{x\to\infty}\frac{\ln x}{x}\overset{\text{L'H}}{=}\lim_{x\to\infty}\frac{1/x}{1}=0$$

differentiate top and bottom; the form was $\infty/\infty$, so this is legal

$$\Rightarrow\ \lim_{n\to\infty}\frac{\ln n}{n}=0$$

the function theorem carries the value $0$ down to the integers

Answer $$\boxed{0}$$
Check

Independent check: $\ln n$ grows slower than any positive power of $n$, so a plain $n$ underneath wins easily.

The one subtlety is the very first line: you differentiate in $x$, never in $n$, because $n$ is a discrete label.

Checkpoint
§01.3 — a first-degree ratio●●○○○

One clean computation to lock in the divide-by-the-top-power reflex. The sequence is a ratio of two first-degree expressions in $n$.

Given

  • $a_n=\dfrac{4n-1}{2n+3}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{4n-1}{2n+3}$.

Hint 1/4

The highest power of $n$ present in numerator and denominator is $n^1$.

Hint 2/4

Divide every term by $n$; the constants become $1/n$ pieces that vanish.

Hint 3/4

This gives $\dfrac{4-1/n}{2+3/n}$, and $1/n\to 0$.

Hint 4/4

So the limit is $4/2=2$.

Show solution
Divide by $n$ and drop the vanishing terms
$$\frac{4n-1}{2n+3}=\frac{4-\frac1n}{2+\frac3n}\to\frac{4-0}{2+0}=2$$

each $1/n$ term tends to $0$; only the leading coefficients $4$ and $2$ remain

Answer $$\boxed{2}$$
Check

At $n=1000$, $(3999)/(2003)=1.9965\ldots$, closing on $2$ from below.

⚠ Differentiating in $n$ to apply L'Hopital

the form is $\infty/\infty$, so the rule feels like it should fire directly on $a_n$

wrong$$\frac{d}{dn}\Big(\frac{\ln n}{n}\Big)\ \text{(there is no derivative in a discrete }n)$$
right$$\text{set }f(x)=\tfrac{\ln x}{x},\ \text{differentiate in }x,\ \text{then }a_n=f(n)$$
⚠ Dividing by the wrong power of $n$

grabbing the first power you see instead of the highest one present

wrong$$\frac{3n^2+2n}{n^2-1}\ \text{divided by }n:\ \frac{3n+2}{n-1/n}\to\infty$$
right$$\text{divide by }n^2:\ \frac{3+2/n}{1-1/n^2}\to 3$$

1.4Trapping a wobble: squeeze, absolute value, continuous functions

Handles terms with a bounded oscillating part by trapping them; a magnitude going to zero forces the sequence to zero.

The limit laws need a limit to exist before they fire. For a term carrying a bounded wobble such as $\cos n$ or $(-1)^n$, we do not compute it directly; we trap it between two sequences we can compute.

TheoremTheorem: The Squeeze Theorem for sequences
Conditions

  • the inequality $b_n\le a_n\le c_n$ holds for all $n$ beyond some point

  • the two outer sequences share one limit, $\lim b_n=\lim c_n=L$

  • consequence, absolute-value form: if $\lim|a_n|=0$ then $\lim a_n=0$

$$\boxed{\;b_n\le a_n\le c_n\ \text{and}\ \lim_{n\to\infty}b_n=\lim_{n\to\infty}c_n=L\ \Longrightarrow\ \lim_{n\to\infty}a_n=L\;}$$

If a sequence is pinned between two others that close in on the same number, it has nowhere else to go and closes in on that number too. A special case: since $-|a_n|\le a_n\le|a_n|$, a magnitude going to $0$ drags the sequence to $0$.

Looks like this, but is not

This looks squeezable: $-1\le(-1)^n\le 1$, a clean sandwich with constant bounds.

But the bounds must share one limit: here $\lim(-1)=-1$ and $\lim 1=1$ disagree, so the theorem says nothing, and indeed $(-1)^n$ diverges. The squeeze only concludes a limit when the top and bottom close in on the same number.

Limit of $\dfrac{\cos n}{n}$

Compute $\displaystyle\lim_{n\to\infty}\frac{\cos n}{n}$.

Given

  • $a_n=\dfrac{\cos n}{n}$

Find

The limit.

Solution
Bound the wobble, then divide
$$-1\le\cos n\le 1$$

the cosine of any real number lives in $[-1,1]$; this is the only fact we use about $\cos n$

$$-\frac1n\le\frac{\cos n}{n}\le\frac1n$$

divide the bound by the positive number $n$, which preserves the inequalities

Squeeze
$$\lim_{n\to\infty}\Big(-\frac1n\Big)=\lim_{n\to\infty}\frac1n=0$$

both envelopes share the limit $0$, so the Squeeze Theorem applies

$$\Rightarrow\ \lim_{n\to\infty}\frac{\cos n}{n}=0$$

the trapped term inherits the common limit

Answer $$\boxed{0}$$
Check

Sanity check on size: $|(\cos n)/n|\le 1/n\to 0$, so the absolute-value test gives $0$ by itself.

Any bounded factor over something growing dies: replace $\cos n$ by $\sin n$, $(-1)^n$, or any $|w_n|\le M$.

Limit of $\dfrac{n!}{n^{\,n}}$

Compute $\displaystyle\lim_{n\to\infty}\frac{n!}{n^{n}}$.

Given

  • $a_n=\dfrac{n!}{n^{n}}$

Find

The limit.

Solution
Write the ratio factor by factor
$$\frac{n!}{n^{n}}=\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\frac{n}{n}$$

$n!$ has $n$ factors and so does $n^n$; pair them up

Keep the first factor, bound the rest by 1
$$0<\frac{n!}{n^{n}}\le\frac{1}{n}\cdot 1\cdot 1\cdots 1=\frac1n$$

every later factor $k/n\le 1$, so dropping them to $1$ can only increase the product

$$\lim_{n\to\infty}0=\lim_{n\to\infty}\frac1n=0\ \Rightarrow\ \lim_{n\to\infty}\frac{n!}{n^{n}}=0$$

squeeze between $0$ and $1/n$

Answer $$\boxed{0}$$
Check

Numeric check: at $n=5$, $120/3125=0.0384\le 1/5=0.2$; at $n=10$, about $3.6\times10^{-4}\le 0.1$.

The whole trick is refusing to expand $n!$; one factor kept and the rest bounded by $1$ is enough.

A continuous function passes through the limit

Compute $\displaystyle\lim_{n\to\infty}\sqrt{\frac{n+1}{n}}$.

Given

  • $a_n=\sqrt{\dfrac{n+1}{n}}$

Find

The limit.

Solution
Find the inside limit, then apply the outer function
$$\frac{n+1}{n}=1+\frac1n\to 1$$

the inside is a rational sequence tending to $1$

$$\sqrt{\ }\ \text{is continuous at }1\ \Rightarrow\ \sqrt{\frac{n+1}{n}}\to\sqrt1=1$$

a continuous function may be pulled through the limit: $\lim f(a_n)=f(\lim a_n)$

Answer $$\boxed{1}$$
Check

Direct check: $\sqrt{(n+1)/n}=\sqrt{1+1/n}$, and $1+1/n\to 1$ with the root of $1$ being $1$.

Checkpoint
§01.4 — the absolute-value test●●○○○

Thirty seconds on the shortcut hiding inside the squeeze. When a sequence changes sign but its size goes to $0$, you do not need two envelopes; the magnitude alone settles it.

Given

  • $a_n=\dfrac{(-1)^n}{\sqrt{n}}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{(-1)^n}{\sqrt n}$, and name the tool.

Hint 1/4

The sign flips every step, so no single side is approached; look at the size instead.

Hint 2/4

Absolute-value test: if $|a_n|\to 0$ then $a_n\to 0$, sign notwithstanding.

Hint 3/4

Here $|a_n|=\dfrac{1}{\sqrt n}$, and $1/\sqrt n\to 0$.

Hint 4/4

So $a_n\to 0$ by the absolute-value test.

Show solution
Reduce to the magnitude
$$|a_n|=\frac{1}{\sqrt n}\to 0$$

the alternating sign disappears under the absolute value

$$|a_n|\to 0\ \Rightarrow\ a_n\to 0$$

the absolute-value form of the squeeze finishes it

Answer $$\boxed{0}$$
Check

Envelopes agree: $-1/\sqrt n\le(-1)^n/\sqrt n\le 1/\sqrt n$, both ends $\to 0$.

⚠ Squeezing with bounds that head to different limits

any true inequality looks like a valid sandwich

wrong$$-1\le(-1)^n\le 1\ \Rightarrow\ (-1)^n\to\ ?$$
right$$\text{squeeze needs }\lim b_n=\lim c_n;\ \text{here }-1\ne 1,\ \text{so no conclusion}$$
⚠ Using the absolute-value test for a nonzero limit

the test is remembered as magnitude limit equals the limit

wrong$$|(-1)^n\tfrac{n}{n+1}|\to 1\ \Rightarrow\ (-1)^n\tfrac{n}{n+1}\to 1$$
right$$\text{the test only gives }a_n\to 0\ \text{from }|a_n|\to 0;\ \text{otherwise the sign matters}$$

1.5The geometric sequence and the standard limits

Settles $r^{\,n}$ by comparing $|r|$ with $1$, and lists the standard limits to quote on sight.

One family, $r^{\,n}$, turns up more than any other, and a short table of standard limits settles most of the rest on sight.

TheoremTheorem: the limit of $r^{\,n}$
Conditions

  • $|r|<1$: the powers shrink, $\lim r^{\,n}=0$

  • $r=1$: the constant sequence $1,1,1,\ldots$, $\lim r^{\,n}=1$

  • $r=-1$ or $|r|>1$: the sequence diverges (oscillation or unbounded growth)

$$\boxed{\;\lim_{n\to\infty}r^{\,n}=\begin{cases}0,&|r|<1\\[2pt]1,&r=1\\[2pt]\text{diverges},&r=-1\ \text{or}\ |r|>1\end{cases}\;}$$

Raising a fraction of size below one to higher powers drives it to zero; exactly one keeps it fixed at one; anything of size above one, or the value negative one, refuses to settle. So $r^{\,n}$ converges precisely when $-1<r\le 1$.

Looks like this, but is not

This looks like it must oscillate forever: $(-0.9)^n$ flips sign every step, just like $(-1)^n$.

The sign is harmless once the size shrinks: $|(-0.9)^n|=0.9^n\to 0$, so by the absolute-value test $(-0.9)^n\to 0$. What decides $r^n$ is $|r|$ against $1$; the alternating sign only matters at the boundary $r=-1$, where the size stops shrinking.

sequencelimitwhy it holds

$\dfrac{\ln n}{n}$

$0$

L'Hopital on $(\ln x)/x$; a log loses to any power

$n^{1/n}$

$1$

$=e^{(\ln n)/n}$ and the exponent $\to 0$

$x^{1/n}\ (x>0)$

$1$

the $n$-th root pulls any fixed positive number to $1$

$r^{\,n}\ (|r|<1)$

$0$

the geometric theorem above

$\Big(1+\dfrac{x}{n}\Big)^{n}$

$e^{x}$

the defining limit of $e$, with $n$ in place of the real variable

$\dfrac{x^{n}}{n!}$

$0$

the factorial eventually outgrows any fixed power's base

Read these as a lookup table: if a sequence matches a row, quote the limit; if it is a ratio of powers, factor it into an $r^{\,n}$ first.

Two geometric sequences, one each side of the line

Decide $\displaystyle\lim_{n\to\infty}\Big(\tfrac23\Big)^{n}$ and $\displaystyle\lim_{n\to\infty}\Big(\tfrac54\Big)^{n}$.

Given

  • $a_n=(2/3)^n$

  • $b_n=(5/4)^n$

Find

Each limit, or divergence.

Solution
Compare each base with 1
$$|\tfrac23|=\tfrac23<1\ \Rightarrow\ (\tfrac23)^n\to 0$$

size below one, so the powers shrink to zero

$$\tfrac54>1\ \Rightarrow\ (\tfrac54)^n\to\infty,\ \text{diverges}$$

size above one, so the powers grow without bound

Answer $$\boxed{(2/3)^n\to 0,\qquad (5/4)^n\ \text{diverges}}$$
Check

Terms confirm it: $(2/3)^{10}=0.017$ and falling, while $(5/4)^{10}=9.3$ and climbing.

For $r^n$ the whole decision is $|r|$ versus $1$; nothing else in the expression matters.

Limit of $\dfrac{3^{\,n+1}}{4^{\,n}}$

Compute $\displaystyle\lim_{n\to\infty}\frac{3^{\,n+1}}{4^{\,n}}$.

Given

  • $a_n=\dfrac{3^{\,n+1}}{4^{\,n}}$

Find

The limit.

Solution
Factor into a constant times a geometric sequence
$$\frac{3^{\,n+1}}{4^{\,n}}=3\cdot\frac{3^{\,n}}{4^{\,n}}=3\Big(\tfrac34\Big)^{n}$$

split $3^{n+1}=3\cdot 3^n$, then group the powers into one base $3/4$

Read off the geometric limit
$$\tfrac34<1\ \Rightarrow\ \Big(\tfrac34\Big)^{n}\to 0$$

size below one

$$3\cdot 0=0$$

a constant times a sequence going to zero still goes to zero

Answer $$\boxed{0}$$
Check

Check the trend: $3(3/4)^n$ gives $2.25,1.69,1.27,\ldots$, steadily halving-ish toward $0$.

The move that unlocks any ratio of exponentials is folding it into a single $r^n$.

Limit of $n^{1/n}$

Compute $\displaystyle\lim_{n\to\infty} n^{1/n}$.

Given

  • $a_n=n^{1/n}$

Find

The limit.

Solution
Rewrite the root as an exponential
$$n^{1/n}=e^{\frac{1}{n}\ln n}=e^{(\ln n)/n}$$

$a^{b}=e^{b\ln a}$; this moves the awkward exponent into a limit we know

Use the standard limit and continuity of $e^{x}$
$$\frac{\ln n}{n}\to 0\ \Rightarrow\ e^{(\ln n)/n}\to e^{0}=1$$

the exponent tends to $0$, and $e^{x}$ is continuous, so pass the limit inside

Answer $$\boxed{1}$$
Check

Values rise then fall to $1$: $2^{1/2}=1.41$, $3^{1/3}=1.44$, $10^{1/10}=1.26$, $1000^{1/1000}=1.007$.

Any polynomial in $n$ raised to $1/n$ also tends to $1$, for the same exponent-to-zero reason.

Checkpoint
§01.5 — a ratio that hides a geometric sequence●●●○○

Thirty seconds on a ratio of exponentials that looks like it needs work but collapses once you spot the $(3/4)^n$ inside it.

Given

  • $a_n=\dfrac{4^{\,n}}{4^{\,n}+3^{\,n}}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{4^{\,n}}{4^{\,n}+3^{\,n}}$.

Hint 1/4

The biggest base present is $4$; make it the yardstick everything is measured against.

Hint 2/4

Divide numerator and denominator by $4^{\,n}$, and recognise $3^{\,n}/4^{\,n}=(3/4)^{n}$.

Hint 3/4

This gives $\dfrac{1}{1+(3/4)^{n}}$, and $(3/4)^{n}\to 0$ since $3/4<1$.

Hint 4/4

So the limit is $\dfrac{1}{1+0}=1$.

Show solution
Normalise by the largest base
$$\frac{4^{n}}{4^{n}+3^{n}}=\frac{1}{1+(3/4)^{n}}$$

divide top and bottom by $4^n$; the cross term becomes a geometric sequence

$$(3/4)^{n}\to 0\ \Rightarrow\ \frac{1}{1+0}=1$$

the base $3/4$ is below one, so its powers vanish

Answer $$\boxed{1}$$
Check

At $n=5$: $1024/(1024+243)=0.808$; at $n=10$: about $0.947$, climbing to $1$.

⚠ Thinking $(-1)^n$ converges because its size is fixed

the boundary $r=-1$ looks as tame as $r=1$

wrong$$r=-1:\ (-1)^n\to 1\ \text{(size is 1)}$$
right$$r=-1:\ (-1)^n\ \text{diverges; the size never shrinks and the sign never settles}$$
⚠ Calling $r^{\,n}$ convergent for $r>1$

writing a limit symbol makes any expression feel convergent

wrong$$\lim(1.1)^n=\infty\ \text{so it converges to }\infty$$
right$$\lim(1.1)^n=\infty\ \text{is a divergence; there is no finite limit}$$

1.6Recursions: the monotonic sequence theorem

Certifies that a bounded monotonic sequence converges, so recursions with no closed form still have a findable limit.

For a recursion $a_{n+1}=f(a_n)$ there is usually no formula for $a_n$, so no limit law can reach it. The Monotonic Sequence Theorem certifies convergence from shape alone: keep rising, stay under a ceiling, and you must arrive somewhere.

TheoremTheorem: The Monotonic Sequence Theorem
Conditions

  • monotonic: either $a_{n+1}\ge a_n$ for all $n$ (increasing) or $a_{n+1}\le a_n$ for all $n$ (decreasing)

  • bounded: increasing needs a ceiling $a_n\le M$; decreasing needs a floor $a_n\ge m$

  • the theorem gives existence of the limit, not its value; you still solve for the value separately

$$\boxed{\;\{a_n\}\ \text{monotonic and bounded}\ \Longrightarrow\ \{a_n\}\ \text{converges}\;}$$

A sequence that only ever climbs but can never pass a fixed ceiling has to pile up against some height at or below that ceiling, and that height is its limit. The mirror statement holds for a sequence that only ever falls but stays above a floor.

Looks like this, but is not

This looks like the theorem should apply: $a_n=n$ is increasing, marching up $1,2,3,\ldots$ without ever turning back.

Monotonic is only half the hypothesis: $a_n=n$ has no ceiling, so it is unbounded and diverges to $\infty$. You need both monotonic and bounded; increasing alone lets a sequence run off to infinity.

The dosing sequence converges to $300$ mg

For $a_1=150$, $a_{n+1}=\tfrac12 a_n+150$, show the sequence converges and find its limit.

Given

  • $a_1=150$

  • $a_{n+1}=\tfrac12 a_n+150$

Find

A proof of convergence and the limit value.

Solution
Prove it stays under 300 (bounded above)
$$a_1=150<300$$

the base case of an induction: the first term is under the ceiling

$$a_n<300\ \Rightarrow\ a_{n+1}=\tfrac12 a_n+150<\tfrac12(300)+150=300$$

if a term is under $300$, the rule keeps the next one under $300$ too

Prove it keeps rising (increasing)
$$a_{n+1}-a_n=\tfrac12 a_n+150-a_n=150-\tfrac12 a_n$$

subtract consecutive terms to see the direction of change

$$a_n<300\ \Rightarrow\ 150-\tfrac12 a_n>150-150=0$$

since every term is under $300$, the difference is positive, so $a_{n+1}>a_n$

Invoke the theorem, then solve for the limit
$$\text{increasing}+\text{bounded above}\ \Rightarrow\ \lim a_n=L\ \text{exists}$$

the Monotonic Sequence Theorem supplies existence; only now may we name $L$

$$L=\tfrac12 L+150\ \Rightarrow\ \tfrac12 L=150\ \Rightarrow\ L=300$$

take the limit of both sides of the recursion; $a_{n+1}$ and $a_n$ share the limit $L$

Answer $$\boxed{\lim_{n\to\infty}a_n=300\ \text{mg}}$$
Check

Both hypotheses were used: without the ceiling, $L=300$ would be an unjustified guess. Terms $290.6,295.3,297.7,\ldots$ approach $300$ from below.

This is the hook resolved: the drug levels off at exactly twice a dose, never accumulating past $300$ mg.

Limit of $a_1=1$, $a_{n+1}=\sqrt{2+a_n}$

The sequence $a_1=1$, $a_{n+1}=\sqrt{2+a_n}$ is increasing and bounded above by $2$. Find its limit.

Given

  • $a_1=1$

  • $a_{n+1}=\sqrt{2+a_n}$

  • given: increasing and $a_n<2$

Find

The limit value.

Solution
Use the given convergence to set up the fixed point
$$\text{bounded}+\text{increasing}\ \Rightarrow\ \lim a_n=L\ \text{exists}$$

the Monotonic Sequence Theorem, from the two facts we are given, licenses naming $L$

$$L=\sqrt{2+L}$$

both sides of $a_{n+1}=\sqrt{2+a_n}$ have limit $L$

Solve, and reject the impossible root
$$L^2=2+L\ \Rightarrow\ L^2-L-2=0\ \Rightarrow\ (L-2)(L+1)=0$$

square to clear the root, then factor the quadratic

$$L=2\ \text{or}\ L=-1;\ \text{take }L=2$$

the terms are positive, so $L\ge 0$; the root $-1$ is discarded

Answer $$\boxed{L=2}$$
Check

The rejected root $-1$ fails the original equation ($\sqrt{2+(-1)}=1\ne -1$); only $L=2$ is consistent.

Squaring can invent roots, so the sign check that kills $-1$ is part of the method, not an afterthought.

Checkpoint
§01.6 — the fixed point of a recursion●●○○○

Thirty seconds on the last step of every convergent recursion. You are told the sequence $a_1=2$, $a_{n+1}=\tfrac12 a_n+3$ converges, so the fixed-point equation is now legal to solve.

Given

  • $a_{n+1}=\tfrac12 a_n+3$

  • given: the sequence converges

Find

  1. (a) Find the limit $L$.

Hint 1/4

Convergence is given, so both $a_{n+1}$ and $a_n$ share one limit $L$; the recursion becomes an equation in $L$.

Hint 2/4

Replace $a_{n+1}$ and $a_n$ by $L$: the rule $a_{n+1}=\tfrac12 a_n+3$ turns into $L=\tfrac12 L+3$.

Hint 3/4

Solve $L=\tfrac12 L+3$: subtract $\tfrac12 L$ to get $\tfrac12 L=3$.

Hint 4/4

So $L=6$.

Show solution
Pass to the limit and solve
$$L=\tfrac12 L+3$$

both sides of the recursion have the same limit $L$ because convergence is given

$$\tfrac12 L=3\ \Rightarrow\ L=6$$

collect the $L$ terms and solve the linear equation

Answer $$\boxed{6}$$
Check

Check the fixed point: $\tfrac12(6)+3=6$, so $6$ maps to itself under the rule, as a limit must.

⚠ Solving $L=f(L)$ without first proving convergence

the fixed-point equation is easy and feels like the whole job

wrong$$a_{n+1}=2a_n,\ a_1=1:\ L=2L\Rightarrow L=0\ \text{(but the sequence is }1,2,4,\ldots\to\infty)$$
right$$\text{first secure convergence; a divergent recursion also solves }L=f(L)$$
⚠ Reading increasing as convergent

an always-rising sequence feels like it is heading somewhere definite

wrong$$a_n=n\ \text{increasing}\ \Rightarrow\ \text{converges}$$
right$$a_n=n\ \text{increasing but unbounded}\ \Rightarrow\ \text{diverges; need a ceiling too}$$
Finding the limit of a sequence: the order of attack

Any question that asks for $\lim a_n$ or whether $\{a_n\}$ converges.

  1. Strip the sign

    If a factor $(-1)^n$ is present, work with the magnitude $|a_n|$ first. If $|a_n|\to 0$, then $a_n\to 0$; if $|a_n|\to L>0$ while the sign alternates, the sequence diverges.

  2. Ratio of powers

    For a ratio of polynomials, divide by the highest power of $n$ and drop the vanishing terms: the answer is the ratio of leading coefficients (or $0$/diverges if the degrees differ).

  3. Ratio of exponentials

    Fold powers into a single base: $a^n/b^n=(a/b)^n$. Then $r^n\to 0$ if $|r|<1$, $\to 1$ if $r=1$, and diverges otherwise.

  4. Stuck algebraically

    Pass to a function $f(x)$ with $a_n=f(n)$ and use L'Hopital or a known function limit; or recognise a standard limit such as $n^{1/n}\to 1$ or $(1+x/n)^n\to e^x$.

  5. Bounded wobble

    For $\cos n$, $\sin n$ or any $|w_n|\le M$ over something growing, squeeze between $\pm M/(\text{growing})$.

  6. Recursion

    No formula for $a_n$? Prove monotonic and bounded, invoke the Monotonic Sequence Theorem, then solve $L=f(L)$.

Where it goes wrong

  • Solving $L=f(L)$ before convergence is proved.

  • Dividing by the wrong power of $n$.

  • Forgetting that $r=-1$ diverges.

Proving a recursion converges (and finding the limit)

A sequence given by $a_{n+1}=f(a_n)$ and a starting value, with no closed form.

  1. Guess the bound

    Compute a few terms; guess a ceiling $M$ (if rising) or floor (if falling). The fixed point of $L=f(L)$ is the natural candidate.

  2. Prove bounded by induction

    Base case: check the first term obeys the bound. Step: assume $a_n$ obeys it and show $a_{n+1}=f(a_n)$ does too.

  3. Prove monotonic

    Show the sign of $a_{n+1}-a_n$ is constant, usually by using the bound just proved.

  4. Invoke the theorem

    Bounded and monotonic gives a finite limit $L$ by the Monotonic Sequence Theorem. Only now is $L$ known to exist.

  5. Solve the fixed point

    Take limits in $a_{n+1}=f(a_n)$ to get $L=f(L)$; solve, and reject roots that violate the sign or the bound.

Where it goes wrong

  • Naming $L$ before step 4.

  • Keeping an extraneous root from squaring.

$\dfrac{(-1)^n}{n}$ converges to $0$

Decide $\lim (-1)^n/n$.

Given

  • $a_n=(-1)^n/n$

Find

The limit.

Solution
Look at the magnitude
$$|a_n|=\tfrac1n\to 0$$

the size shrinks to zero

$$|a_n|\to 0\Rightarrow a_n\to 0$$

absolute-value test: a vanishing size drags the term to $0$ whatever the sign

Answer $$\boxed{0}$$
Check

Terms $-1,\tfrac12,-\tfrac13,\tfrac14$ close in on $0$ from both sides.

$(-1)^n\dfrac{n}{n+1}$ diverges

Decide $\lim (-1)^n n/(n+1)$.

Given

  • $b_n=(-1)^n n/(n+1)$

Find

The limit, or divergence.

Solution
Look at the magnitude, then the sign
$$|b_n|=\tfrac{n}{n+1}\to 1$$

the size settles on $1$, not $0$

$$\text{even }n\to +1,\ \text{odd }n\to -1$$

with the size fixed near $1$, the sign splits the terms toward $+1$ and $-1$

Answer $$\boxed{\text{diverges}}$$
Check

Even terms $\tfrac23,\tfrac45,\ldots\to 1$; odd terms $-\tfrac12,-\tfrac34,\ldots\to -1$; two limits, so none.

Both carry the same alternating sign $(-1)^n$; the only difference is whether the magnitude tends to $0$ or to $1$, and that single fact decides convergence.

How to tell them apart

An alternating sign is harmless exactly when the magnitude tends to $0$. If the magnitude tends to a positive number, the sign makes the sequence diverge.

Scaffolding comes off
The common skeleton

  1. Strip any $(-1)^n$ and look at the magnitude $|a_n|$ on its own.

  2. Find the limit of the magnitude: for a ratio of powers, divide by the highest power of $n$.

  3. If the magnitude tends to $0$, the sequence tends to $0$ (the sign cannot save a shrinking size).

  4. If the magnitude tends to $L>0$: a fixed sign gives limit $\pm L$; an alternating sign splits even and odd, so it diverges.

  5. Confirm with one even-index and one odd-index term.

1 · fully worked

Full worked: $\lim\dfrac{4n+3}{2n-1}$

Compute $\displaystyle\lim_{n\to\infty}\frac{4n+3}{2n-1}$.

Given

  • $a_n=\dfrac{4n+3}{2n-1}$

Find

The limit.

Solution
No sign to strip; go to the magnitude, a ratio of first-degree terms
$$a_n>0\ \text{for all }n,\ \text{so }|a_n|=a_n$$

there is no alternating factor, so the sign step is trivial

Divide by the highest power, $n$
$$\frac{4n+3}{2n-1}=\frac{4+3/n}{2-1/n}$$

dividing top and bottom by $n$ converts growth into vanishing $1/n$ terms

$$\to\frac{4+0}{2-0}=2$$

each $1/n$ tends to $0$; the limit is the ratio of leading coefficients

Confirm
$$a_{100}=\frac{403}{199}=2.0251\ldots$$

a large-$n$ value sits just above $2$, as a $1/n$-sized error predicts

Answer $$\boxed{2}$$
Check

Both an even index ($a_{100}=2.025$) and an odd one ($a_{101}=402... /201=2.0248$) hug $2$.

2 · you write the reasoning

Easier, and now you supply the reasons. Compute $$\lim_{n\to\infty}\frac{5n+2}{n}$$ and write, in the empty column, why each line is allowed.

  1. reasoning

    Splitting the fraction term by term is legal for any $n$ and separates the constant part from the vanishing part.

  2. reasoning

    The reciprocal $2/n$ is a constant over a growing denominator, a standard sequence tending to $0$.

  3. reasoning

    By the sum law the limit of $5+2/n$ is $5+0$, since a constant has limit itself and $2/n\to 0$.

3 · find the buried error

Harder, and the work is done for you, badly. Two of the four steps are wrong. Find both. The task: decide whether $$a_n=(-1)^n\frac{n}{2n+1}$$ converges, and if so give the limit.

the two buried errors (2)
⚠ step 2

The absolute-value test was used to turn a magnitude limit into the sequence's limit. That test only concludes $a_n\to 0$ from $|a_n|\to 0$; when the magnitude tends to a positive number, the sign cannot be ignored.

The rule is half-remembered as magnitude limit equals the limit, and the special role of $0$ is dropped.

right

From $|a_n|\to\tfrac12$ you may only say the terms have size near $\tfrac12$; the sign still has to be examined separately.

⚠ step 4

Only even-indexed terms were checked in step 3, and their trend was read as the whole sequence's limit. The odd terms $a_1=-\tfrac13$, $a_3=-\tfrac37$ head to $-\tfrac12$, so even and odd terms have different limits.

Testing a couple of convenient terms and generalising is a natural shortcut, and the alternating sign is easy to forget once a positive number appears.

right

Even terms $\to+\tfrac12$ and odd terms $\to-\tfrac12$: two subsequential limits, so $a_n$ diverges.

4 · the bare problem
§01.6 — decide convergence, bare●●●○○

No scaffolding this time. Use the skeleton: strip the sign, size up the magnitude, then let the sign decide.

Given

  • $b_n=(-1)^n\dfrac{n}{n+4}$

Find

  1. (a) Does $\{b_n\}$ converge? If so, give the limit; if not, say why.

Hint 1/4

Separate the alternating sign from the size; decide the size first.

Hint 2/4

The magnitude is $|b_n|=\dfrac{n}{n+4}$; find its limit by dividing by $n$.

Hint 3/4

Here $|b_n|=\dfrac{1}{1+4/n}\to 1$, a positive number, and the sign alternates.

Hint 4/4

A fixed positive size with alternating sign splits even ($\to +1$) and odd ($\to -1$): it diverges.

Show solution
Size first
$$|b_n|=\frac{n}{n+4}=\frac{1}{1+4/n}\to 1$$

divide by $n$; the magnitude settles on $1$, not $0$

Then the sign
$$\text{even }n:\ b_n\to +1,\qquad \text{odd }n:\ b_n\to -1$$

with size near $1$ and an alternating sign, the two subsequences separate

$$\text{two different limits}\ \Rightarrow\ \text{diverges}$$

a convergent sequence has exactly one limit

Answer $$\boxed{\text{diverges}}$$
Check

Witnesses: $b_{100}=100/104=0.96$ (even, positive) and $b_{101}=-101/105=-0.96$ (odd, negative).

Whenever the magnitude tends to a nonzero number and the sign alternates, stop: it diverges.

Full exam-style question

Determine convergence and the limit of $\dfrac{2n^2-3}{5n^2+n}$exam format

Determine whether $a_n=\dfrac{2n^2-3}{5n^2+n}$ converges, and if so find its limit.

Given

  • $a_n=\dfrac{2n^2-3}{5n^2+n}$

Find

A convergence verdict and, if it converges, the limit.

Solution
Identify the dominant power and divide
$$a_n=\frac{2n^2-3}{5n^2+n}=\frac{2-3/n^2}{5+1/n}$$

numerator and denominator are both degree $2$; divide every term by $n^2$

Send the small terms to zero
$$\frac{2-3/n^2}{5+1/n}\to\frac{2-0}{5+0}=\frac25$$

$3/n^2\to0$ and $1/n\to0$ by the limit laws; the leading coefficients remain

$$\text{a finite limit exists}\ \Rightarrow\ \text{converges}$$

having a single finite limit is exactly convergence

Answer $$\boxed{\text{converges},\quad \lim_{n\to\infty}a_n=\tfrac25}$$
Check

At $n=100$, $a_n=(20000-3)/(50000+100)=19997/50100=0.3992$, within $0.001$ of $0.4=\tfrac25$.

Equal degrees on top and bottom always give a finite nonzero limit: the ratio of leading coefficients.

The exam phrasing wants two things: the word converges and the number; supply both, and show the division that earns them.

Practice

A · concept 4 questions
1§01.6 — bounded does not imply convergent●●○○○

A revision partner claims that once a sequence is bounded it cannot run off anywhere, so it has to converge. Decide the precise version below.

Given

  • Claim: every bounded sequence converges.

Find

  1. (a) True or false, with a supporting sequence.

Hint 1/4

Ask whether bounded rules out the terms from oscillating.

Hint 2/4

Bounded means the terms stay in some interval; it says nothing about settling on one value.

Hint 3/4

Test $(-1)^n$: it stays in $[-1,1]$ (bounded) but has no limit.

Hint 4/4

So the claim is false; $(-1)^n$ is the standard counterexample.

Show solution
Exhibit a bounded divergent sequence
$$(-1)^n\in[-1,1]\ \text{for all }n$$

the terms never leave a fixed interval, so the sequence is bounded

$$\text{even}\to+1,\ \text{odd}\to-1\ \Rightarrow\ \text{diverges}$$

two clustering values means no single limit

Answer $$\text{False}$$
Check

Boundedness plus monotonicity would force convergence; boundedness alone does not.

2§01.4 — a magnitude limit is not a limit●●○○○

The absolute-value test is easy to over-read. A classmate writes: if the sizes $|a_n|$ settle down, then the terms $a_n$ settle down too. Decide it.

Given

  • Claim: if $\lim|a_n|$ exists then $\lim a_n$ exists.

Find

  1. (a) True or false, with a supporting sequence.

Hint 1/4

Separate the size of the terms from the terms themselves.

Hint 2/4

The test only says $|a_n|\to 0$ forces $a_n\to 0$; a nonzero magnitude limit says nothing about the sign.

Hint 3/4

Test $(-1)^n$: $|a_n|=1\to 1$ exists, but $a_n$ diverges.

Hint 4/4

So the claim is false; the magnitude can converge while the sequence does not.

Show solution
Use a fixed magnitude with an unsettled sign
$$|(-1)^n|=1\to 1$$

the magnitude is the constant sequence $1$, which converges

$$(-1)^n\ \text{diverges}$$

the sign never settles, so the sequence has no limit

Answer $$\text{False}$$
Check

Only the $0$ case is safe: $|a_n|\to 0$ does force $a_n\to 0$.

3§01.5 — spot the divergent one●●○○○

Four sequences are listed; three settle to a finite value and one does not. The single new idea being tested is the boundary $|r|$ against $1$ for a geometric sequence.

Given

  • $\Big(\tfrac{9}{10}\Big)^{n},\quad \tfrac{1}{n},\quad \Big(\tfrac{5}{4}\Big)^{n},\quad \tfrac{n}{n+1}$

Find

  1. (a) Which one diverges?

Hint 1/4

Three of these clearly settle; scan for the one whose terms grow without bound.

Hint 2/4

A geometric $r^n$ diverges when $|r|>1$; check each base against $1$.

Hint 3/4

Here $9/10<1$ (to $0$), $1/n\to0$, $n/(n+1)\to1$, but $5/4>1$.

Hint 4/4

So $(5/4)^n$ diverges; the other three converge.

Show solution
Classify each
$$(9/10)^n\to0,\quad 1/n\to0,\quad n/(n+1)\to1$$

two geometric-or-rational sequences to $0$ and one rational to $1$, all finite

$$(5/4)^n:\ 5/4>1\ \Rightarrow\ \to\infty$$

size above one, so the powers grow without bound

Answer $$\boxed{(5/4)^n}$$
Check

Numerically $(5/4)^{20}\approx86.7$ and rising, unlike the other three.

4§01.2 — a sequence is not the set of its values●●○○○

One more concept trap that fuses two ideas. A classmate argues that since $(-1)^n$ only ever produces the two numbers $-1$ and $1$, it is really the set $\{-1,1\}$ and therefore converges to one of them.

Given

  • Claim: $(-1)^n$ equals the set $\{-1,1\}$, so it converges.

Find

  1. (a) True or false, and say where the argument breaks.

Hint 1/4

A sequence keeps order and repetition; a set forgets both.

Hint 2/4

Convergence is a statement about the ordered tail of terms, not about the collection of values.

Hint 3/4

The ordered list $-1,1,-1,1,\ldots$ keeps flipping, so its even and odd parts head to $+1$ and $-1$.

Hint 4/4

So the claim is false: the sequence diverges even though its value set has two elements.

Show solution
Separate the two objects
$$\{-1,1\}\ \text{is an unordered set of two values}$$

collapsing repeats and order loses the information a limit needs

$$(-1)^n=-1,1,-1,1,\ldots\ \text{diverges}$$

the ordered tail never settles, so there is no limit

Answer $$\text{False}$$
Check

The value set being finite has no bearing on convergence.

B · computation 7 questions
1§01.3 — quadratic over quadratic●●○○○

A routine ratio to keep the divide-by-top-power reflex sharp. Both numerator and denominator are degree two.

Given

  • $a_n=\dfrac{5n^2-3}{2n^2+7}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{5n^2-3}{2n^2+7}$.

Hint 1/4

The highest power present is $n^2$ in both.

Hint 2/4

Divide every term by $n^2$; constants become $1/n^2$ pieces that vanish.

Hint 3/4

This gives $\dfrac{5-3/n^2}{2+7/n^2}$, and $1/n^2\to0$.

Hint 4/4

So the limit is $5/2$.

Show solution
Divide by $n^2$
$$\frac{5n^2-3}{2n^2+7}=\frac{5-3/n^2}{2+7/n^2}\to\frac{5}{2}$$

both $3/n^2$ and $7/n^2$ vanish; the leading coefficients decide

Answer $$\boxed{5/2}$$
Check

At $n=100$: $(49997)/(20007)=2.4990\ldots$, near $2.5$.

2§01.4 — bounded wobble over n●●○○○

A sine on top makes the numerator wobble but never grow. That is the classic setting for a squeeze.

Given

  • $a_n=\dfrac{\sin n}{n}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{\sin n}{n}$.

Hint 1/4

The numerator is bounded; only its size, not its behaviour, matters.

Hint 2/4

Use $-1\le\sin n\le 1$, then divide by the positive $n$.

Hint 3/4

This traps $\dfrac{\sin n}{n}$ between $-1/n$ and $1/n$, both $\to 0$.

Hint 4/4

By the Squeeze Theorem the limit is $0$.

Show solution
Bound and squeeze
$$-\tfrac1n\le\tfrac{\sin n}{n}\le\tfrac1n$$

from $|\sin n|\le 1$, dividing by $n>0$

$$\pm\tfrac1n\to0\ \Rightarrow\ \tfrac{\sin n}{n}\to0$$

both envelopes share the limit $0$

Answer $$\boxed{0}$$
Check

Equivalently $|(\sin n)/n|\le1/n\to0$, so the absolute-value test gives $0$.

3§01.5 — ratio of two exponentials●○○○○

A quotient of powers that folds into a single geometric sequence once you group the bases.

Given

  • $a_n=\dfrac{2^{\,n}}{5^{\,n}}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{2^{\,n}}{5^{\,n}}$.

Hint 1/4

Two powers with the same exponent can be merged into one base.

Hint 2/4

Write $\dfrac{2^n}{5^n}=\Big(\dfrac25\Big)^{n}$ and compare $2/5$ with $1$.

Hint 3/4

Since $2/5<1$, the geometric sequence $(2/5)^n\to 0$.

Hint 4/4

So the limit is $0$.

Show solution
Group and read off
$$\frac{2^n}{5^n}=\Big(\tfrac25\Big)^n,\quad \tfrac25<1\ \Rightarrow\ \to0$$

a single geometric base below one, so the powers vanish

Answer $$\boxed{0}$$
Check

$(2/5)^5=0.0102$, already tiny and shrinking.

4§01.3 — a negative leading ratio●●○○○

The same divide-by-top-power move, but the leading coefficients carry signs, so watch them.

Given

  • $a_n=\dfrac{4n^2-n}{1-2n^2}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{4n^2-n}{1-2n^2}$.

Hint 1/4

Highest power is $n^2$; the denominator's leading term is $-2n^2$.

Hint 2/4

Divide by $n^2$ and keep the sign of each leading coefficient.

Hint 3/4

This gives $\dfrac{4-1/n}{1/n^2-2}$, tending to $\dfrac{4}{-2}$.

Hint 4/4

So the limit is $-2$.

Show solution
Divide by $n^2$
$$\frac{4n^2-n}{1-2n^2}=\frac{4-1/n}{1/n^2-2}\to\frac{4}{-2}=-2$$

the vanishing terms drop; the denominator's leading coefficient is $-2$

Answer $$\boxed{-2}$$
Check

At $n=100$: $(39900)/(-19999)=-1.9951\ldots$, near $-2$.

5§01.3 — a power beats a polynomial●●●○○

No algebra simplifies $n/2^n$, so pass to a function and let L'Hopital decide which growth wins.

Given

  • $a_n=\dfrac{n}{2^{\,n}}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{n}{2^{\,n}}$.

Hint 1/4

There is nothing to cancel; move to $f(x)=x/2^x$ and check the form $\infty/\infty$.

Hint 2/4

L'Hopital: differentiate top and bottom, using $\dfrac{d}{dx}2^x=2^x\ln 2$.

Hint 3/4

This gives $\dfrac{1}{2^x\ln 2}$, and $2^x\to\infty$.

Hint 4/4

So the limit is $0$: the exponential outgrows the linear term.

Show solution
Function then L'Hopital
$$f(x)=\frac{x}{2^{x}},\ \frac{\infty}{\infty}\overset{\text{L'H}}{=}\frac{1}{2^{x}\ln 2}$$

differentiate in $x$; the denominator's derivative is $2^x\ln2$

$$2^{x}\to\infty\ \Rightarrow\ \frac{1}{2^{x}\ln2}\to0$$

a bounded numerator over an exploding denominator

Answer $$\boxed{0}$$
Check

Terms $0.5,0.5,0.375,0.25,0.156,\ldots$ fall to $0$; any $r^n$ with $r>1$ beats any power of $n$.

6§01.4 — alternating over a growing denominator●●○○○

An alternating numerator over something growing: the sign flips, the size shrinks, so the absolute-value test settles it in one line.

Given

  • $a_n=\dfrac{(-1)^n}{n^2+1}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{(-1)^n}{n^2+1}$.

Hint 1/4

Sign flips every term, so look at the size instead of a side.

Hint 2/4

Absolute-value test: if $|a_n|\to0$ then $a_n\to0$.

Hint 3/4

Here $|a_n|=\dfrac{1}{n^2+1}\to0$.

Hint 4/4

So the limit is $0$.

Show solution
Magnitude to zero
$$|a_n|=\frac{1}{n^2+1}\to0\ \Rightarrow\ a_n\to0$$

the size vanishes, so the alternating sign cannot prevent the limit $0$

Answer $$\boxed{0}$$
Check

Terms $-\tfrac12,\tfrac15,-\tfrac1{10},\ldots$ shrink to $0$ from both sides.

7§01.6 — fixed point of a square-root recursion●●●○○

A recursion that rises toward a ceiling. Take convergence as given (it is increasing and bounded above by $3$) and pin down the value.

Given

  • $a_1=0$

  • $a_{n+1}=\sqrt{6+a_n}$

  • given: increasing and bounded above by $3$

Find

  1. (a) Find the limit $L$.

Hint 1/4

Convergence is given, so $a_{n+1}$ and $a_n$ both tend to the same $L$.

Hint 2/4

Pass to the limit in $a_{n+1}=\sqrt{6+a_n}$ to get $L=\sqrt{6+L}$.

Hint 3/4

Square: $L^2=6+L$, so $L^2-L-6=0$, i.e. $(L-3)(L+2)=0$.

Hint 4/4

The terms are non-negative, so reject $L=-2$; thus $L=3$.

Show solution
Fixed point
$$L=\sqrt{6+L}\ \Rightarrow\ L^2-L-6=0$$

square both sides, valid because $L\ge0$

$$(L-3)(L+2)=0\ \Rightarrow\ L=3$$

reject $L=-2$ since all terms are non-negative

Answer $$\boxed{3}$$
Check

Check: $\sqrt{6+3}=3$, so $3$ is fixed by the rule, as a limit must be.

C · exam level 4 questions
1§01.3 — a root over n●●●○○

Exam-style, one clean idea: a square root of a quadratic behaves like the root of its leading term for large $n$.

Given

  • $a_n=\dfrac{\sqrt{9n^2+2n}}{n}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\frac{\sqrt{9n^2+2n}}{n}$.

Hint 1/4

Bring the $n$ under the root as $n=\sqrt{n^2}$ to compare like with like.

Hint 2/4

Then $\dfrac{\sqrt{9n^2+2n}}{\sqrt{n^2}}=\sqrt{\dfrac{9n^2+2n}{n^2}}=\sqrt{9+2/n}$.

Hint 3/4

As $n\to\infty$, $2/n\to0$, so the inside tends to $9$.

Hint 4/4

So the limit is $\sqrt9=3$.

Show solution
Combine under one root
$$\frac{\sqrt{9n^2+2n}}{n}=\sqrt{9+\tfrac2n}\to\sqrt9=3$$

write $n=\sqrt{n^2}$, divide inside, and let $2/n\to0$; the root is continuous

Answer $$\boxed{3}$$
Check

At $n=100$: $\sqrt{90200}/100=3.0033$, near $3$.

2§01.5 — which one converges●●●○○

Exam-style sorting. Three of these diverge and one converges; the deciding idea is $|r|$ against $1$, with the sign only mattering at the boundary.

Given

  • $(-1)^n,\quad 2^{\,n},\quad (-0.5)^{n},\quad (-1)^n n$

Find

  1. (a) Which sequence converges?

Hint 1/4

Scan for the only one whose terms both shrink and are held near a single value.

Hint 2/4

A geometric $r^n$ converges iff $-1<r\le 1$; check each base.

Hint 3/4

$(-1)^n$ diverges, $2^n\to\infty$, $(-1)^n n$ diverges, but $|-0.5|<1$.

Hint 4/4

So $(-0.5)^n\to0$; it is the convergent one.

Show solution
Classify each
$$(-1)^n,\ (-1)^n n\ \text{diverge};\ 2^n\to\infty$$

oscillation or unbounded growth in three of them

$$|-0.5|<1\ \Rightarrow\ (-0.5)^n\to0$$

size below one, so it converges to $0$ despite the sign

Answer $$\boxed{(-0.5)^n}$$
Check

$|(-0.5)^n|=0.5^n\to0$, so the absolute-value test confirms $0$.

3§01.4 — split off the wobble●●●○○

Exam-style with a short justification expected. The term is a steady part plus an alternating part; separate them.

Given

  • $a_n=\dfrac{n^2+(-1)^n n}{n^2}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}a_n$ and justify each step.

Hint 1/4

Split the single fraction into a part you know and a part that wobbles.

Hint 2/4

Write $a_n=1+\dfrac{(-1)^n}{n}$; the sum law lets you treat the two pieces separately.

Hint 3/4

The wobble $\dfrac{(-1)^n}{n}$ has magnitude $1/n\to0$, so it tends to $0$.

Hint 4/4

So $a_n\to1+0=1$.

Show solution
Separate the pieces
$$a_n=\frac{n^2}{n^2}+\frac{(-1)^n n}{n^2}=1+\frac{(-1)^n}{n}$$

split the fraction; the first piece is exactly $1$

Kill the wobble
$$\Big|\tfrac{(-1)^n}{n}\Big|=\tfrac1n\to0\ \Rightarrow\ \tfrac{(-1)^n}{n}\to0$$

absolute-value test on the alternating piece

$$a_n=1+\tfrac{(-1)^n}{n}\to1$$

sum law: a constant plus a sequence going to $0$

Answer $$\boxed{1}$$
Check

Terms $a_1=0,\ a_2=1.5,\ a_3\approx0.667$ close in on $1$ from alternating sides.

4§01.6 — limit of a nested-root recursion●●●○○

Exam-style. The sequence $a_1=1$, $a_{n+1}=\sqrt{2+a_n}$ can be shown increasing and bounded above by $2$; take that as given and identify the limit.

Given

  • $a_1=1$

  • $a_{n+1}=\sqrt{2+a_n}$

  • given: increasing and bounded above by $2$

Find

  1. (a) What is the limit?

Hint 1/4

Convergence is handed to you, so the fixed-point equation is legal.

Hint 2/4

Set $L=\sqrt{2+L}$ and square to remove the root.

Hint 3/4

This gives $L^2-L-2=0$, i.e. $(L-2)(L+1)=0$.

Hint 4/4

Terms are positive, so reject $-1$; the limit is $2$.

Show solution
Solve the fixed point
$$L=\sqrt{2+L}\ \Rightarrow\ L^2-L-2=0\ \Rightarrow\ (L-2)(L+1)=0$$

square, then factor

$$L=2$$

reject $L=-1$ since the terms are positive

Answer $$\boxed{2}$$
Check

Terms $1,\sqrt3=1.73,\sqrt{3.73}=1.93,\ldots$ climb toward $2$.

D · interleaved 3 questions
1§01.3 — a sequence hiding a Calculus I limit●●●○○

Mixed practice: the type is not announced. A product of a growing factor and a shrinking sine is exactly a limit you met for functions.

Given

  • $a_n=n\sin\!\Big(\dfrac{1}{n}\Big)$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty} n\sin\!\Big(\frac1n\Big)$.

Hint 1/4

It is $\infty\cdot 0$; rewrite it so a familiar limit appears.

Hint 2/4

Let $t=1/n\to0$; then $n\sin(1/n)=\dfrac{\sin t}{t}$.

Hint 3/4

Recall from Calculus I that $\dfrac{\sin t}{t}\to1$ as $t\to0$.

Hint 4/4

So the limit is $1$.

Show solution
Substitute $t=1/n$
$$n\sin\tfrac1n=\frac{\sin(1/n)}{1/n}=\frac{\sin t}{t},\ t=\tfrac1n\to0$$

rewrite so the standard limit is visible

$$\frac{\sin t}{t}\to1$$

the Calculus I limit at $0$

Answer $$\boxed{1}$$
Check

At $n=1000$: $1000\sin(0.001)=0.99999983$, essentially $1$.

2§01.3 — a difference that needs the conjugate●●●●○

Mixed practice. A difference of a root and a linear term is an $\infty-\infty$ form; the Calculus I conjugate trick rescues it.

Given

  • $a_n=\sqrt{n^2+3n}-n$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\big(\sqrt{n^2+3n}-n\big)$.

Hint 1/4

Both pieces run to $\infty$; multiply by the conjugate over itself to expose the gap.

Hint 2/4

Multiply by $\dfrac{\sqrt{n^2+3n}+n}{\sqrt{n^2+3n}+n}$; the numerator becomes $(n^2+3n)-n^2=3n$.

Hint 3/4

You get $\dfrac{3n}{\sqrt{n^2+3n}+n}=\dfrac{3}{\sqrt{1+3/n}+1}$.

Hint 4/4

As $n\to\infty$ the inside root tends to $1$, so the limit is $3/2$.

Show solution
Rationalise
$$\big(\sqrt{n^2+3n}-n\big)\frac{\sqrt{n^2+3n}+n}{\sqrt{n^2+3n}+n}=\frac{3n}{\sqrt{n^2+3n}+n}$$

the conjugate clears the difference of roots in the numerator

Divide by $n$
$$\frac{3n}{\sqrt{n^2+3n}+n}=\frac{3}{\sqrt{1+3/n}+1}\to\frac{3}{1+1}=\frac32$$

pull $n$ inside the root as $\sqrt{n^2}$; the $3/n$ vanishes

Answer $$\boxed{3/2}$$
Check

At $n=1000$: $\sqrt{1003000}-1000=1001.498-1000=1.498$, near $1.5$.

3§01.5 — a limit that builds the number e●●●○○

Mixed practice. A base tending to $1$ raised to a power tending to $\infty$ is the indeterminate form $1^\infty$; one standard limit handles it.

Given

  • $a_n=\Big(1+\dfrac{2}{n}\Big)^{n}$

Find

  1. (a) Find $\displaystyle\lim_{n\to\infty}\Big(1+\frac2n\Big)^{n}$.

Hint 1/4

Do not expand; recognise the shape of the defining limit of $e$.

Hint 2/4

The standard limit is $\Big(1+\dfrac{x}{n}\Big)^{n}\to e^{x}$.

Hint 3/4

Here $x=2$, so the expression tends to $e^{2}$.

Hint 4/4

So the limit is $e^{2}\approx7.389$.

Show solution
Recognise the standard form
$$\Big(1+\tfrac{x}{n}\Big)^{n}\to e^{x},\ x=2\ \Rightarrow\ e^{2}$$

the sequence version of the defining limit of $e$

Answer $$\boxed{e^{2}}$$
Check

At $n=1000$: $(1.002)^{1000}=7.374$, close to $e^2=7.389$.

Mistake ledger (12 entries)
⚠ Treating a sequence as the set of its values

the brace notation $\{a_n\}$ looks exactly like set notation learned earlier

wrong$$\{(-1)^n\}=\{-1,1\}\ \Rightarrow\ \text{two values, so it converges to one of them}$$
right$$\{(-1)^n\}=-1,1,-1,1,\ldots\ \text{(order kept, so it can diverge)}$$
⚠ Thinking a recursion alone defines the sequence

the rule $a_{n+1}=f(a_n)$ looks complete on its own

wrong$$a_{n+1}=\tfrac12 a_n+150\ \Rightarrow\ \text{one sequence}$$
right$$a_{n+1}=\tfrac12 a_n+150,\ a_1=150\ \Rightarrow\ \text{now pinned down}$$
⚠ Concluding a limit from the first several terms

the early terms often march so steadily that the destination looks obvious

wrong$$0.9,0.99,0.999,\ldots\ \Rightarrow\ \text{limit is }1\ \text{(by inspection)}$$
right$$\text{prove the tail enters every band: }|a_n-1|<\varepsilon\ \text{for }n>N$$
⚠ Assuming a bounded sequence must converge

bounded feels like tamed, and tame sequences seem to settle

wrong$$(-1)^n\in[-1,1]\ \Rightarrow\ \text{converges}$$
right$$(-1)^n\ \text{is bounded but diverges; boundedness alone is not enough}$$
⚠ Differentiating in $n$ to apply L'Hopital

the form is $\infty/\infty$, so the rule feels like it should fire directly on $a_n$

wrong$$\frac{d}{dn}\Big(\frac{\ln n}{n}\Big)\ \text{(there is no derivative in a discrete }n)$$
right$$\text{set }f(x)=\tfrac{\ln x}{x},\ \text{differentiate in }x,\ \text{then }a_n=f(n)$$
⚠ Dividing by the wrong power of $n$

grabbing the first power you see instead of the highest one present

wrong$$\frac{3n^2+2n}{n^2-1}\ \text{divided by }n:\ \frac{3n+2}{n-1/n}\to\infty$$
right$$\text{divide by }n^2:\ \frac{3+2/n}{1-1/n^2}\to 3$$
⚠ Squeezing with bounds that head to different limits

any true inequality looks like a valid sandwich

wrong$$-1\le(-1)^n\le 1\ \Rightarrow\ (-1)^n\to\ ?$$
right$$\text{squeeze needs }\lim b_n=\lim c_n;\ \text{here }-1\ne 1,\ \text{so no conclusion}$$
⚠ Using the absolute-value test for a nonzero limit

the test is remembered as magnitude limit equals the limit

wrong$$|(-1)^n\tfrac{n}{n+1}|\to 1\ \Rightarrow\ (-1)^n\tfrac{n}{n+1}\to 1$$
right$$\text{the test only gives }a_n\to 0\ \text{from }|a_n|\to 0;\ \text{otherwise the sign matters}$$
⚠ Thinking $(-1)^n$ converges because its size is fixed

the boundary $r=-1$ looks as tame as $r=1$

wrong$$r=-1:\ (-1)^n\to 1\ \text{(size is 1)}$$
right$$r=-1:\ (-1)^n\ \text{diverges; the size never shrinks and the sign never settles}$$
⚠ Calling $r^{\,n}$ convergent for $r>1$

writing a limit symbol makes any expression feel convergent

wrong$$\lim(1.1)^n=\infty\ \text{so it converges to }\infty$$
right$$\lim(1.1)^n=\infty\ \text{is a divergence; there is no finite limit}$$
⚠ Solving $L=f(L)$ without first proving convergence

the fixed-point equation is easy and feels like the whole job

wrong$$a_{n+1}=2a_n,\ a_1=1:\ L=2L\Rightarrow L=0\ \text{(but the sequence is }1,2,4,\ldots\to\infty)$$
right$$\text{first secure convergence; a divergent recursion also solves }L=f(L)$$
⚠ Reading increasing as convergent

an always-rising sequence feels like it is heading somewhere definite

wrong$$a_n=n\ \text{increasing}\ \Rightarrow\ \text{converges}$$
right$$a_n=n\ \text{increasing but unbounded}\ \Rightarrow\ \text{diverges; need a ceiling too}$$
Formula card
Sequence as a function on the integers
$$\{a_n\},\ a\colon\{1,2,3,\ldots\}\to\mathbb{R}$$

domain is the positive integers; order and repetition matter

Convergence (epsilon-N)
$$\lim_{n\to\infty}a_n=L\iff\forall\varepsilon>0\ \exists N:\ n>N\Rightarrow|a_n-L|<\varepsilon$$

$L$ finite; the tail past $N$ must lie in every band

Sequence read off a function
$$\lim_{x\to\infty}f(x)=L,\ a_n=f(n)\ \Rightarrow\ \lim_{n\to\infty}a_n=L$$

one-way only; the converse can fail

Squeeze Theorem for sequences
$$b_n\le a_n\le c_n,\ \lim b_n=\lim c_n=L\ \Rightarrow\ \lim a_n=L$$

both envelopes share the same limit

Absolute-value test
$$\lim_{n\to\infty}|a_n|=0\ \Rightarrow\ \lim_{n\to\infty}a_n=0$$

only for the limit $0$; a nonzero magnitude limit is not enough

Continuous function through a limit
$$a_n\to L,\ f\ \text{continuous at}\ L\ \Rightarrow\ f(a_n)\to f(L)$$

$f$ continuous at the limit value

Limit of a geometric sequence
$$\lim_{n\to\infty}r^{\,n}=0\ (|r|<1),\ =1\ (r=1),\ \text{diverges otherwise}$$

converges iff $-1<r\le 1$

The e-limit for sequences
$$\lim_{n\to\infty}\Big(1+\frac{x}{n}\Big)^{n}=e^{x}$$

base $\to 1$, power $\to\infty$ (the $1^\infty$ form)

Monotonic Sequence Theorem
$$\text{monotonic}+\text{bounded}\ \Rightarrow\ \text{convergent}$$

increasing needs a ceiling, decreasing needs a floor

Check yourself

Close the page and write, from memory: what convergence means in one sentence, the six moves for finding a limit, the exact rule for $r^{\,n}$ at $r=1$ and $r=-1$, and the two things you must prove before naming the limit of a recursion. Then reopen and compare; the gaps are your reread list, and none of this is scored.

  • Write the first terms of an explicit and a recursive sequence, and say why finitely many terms cannot fix a limit?

    c-sequence-basics

  • State what $\lim a_n=L$ means with a band, and prove $(-1)^n$ diverges?

    c-limit-of-sequence

  • Find the limit of a rational sequence by dividing by the top power, and use L'Hopital through a function?

    c-limit-laws

  • Squeeze a $\cos n$-type term and apply the absolute-value test correctly (only for limit $0$)?

    c-squeeze-and-abs

  • Decide $r^{\,n}$ from $|r|$ against $1$ and quote the standard limits?

    c-geometric-and-key-limits

  • Prove a recursion is monotonic and bounded, then solve $L=f(L)$ and screen the roots?

    c-monotonic-bounded

Glossary (13 terms)
sequencedizi

An ordered unending list of numbers $a_1,a_2,a_3,\ldots$; equivalently a function whose inputs are the positive integers.

termterim

A single entry $a_n$ of a sequence, the output at input $n$.

convergent sequenceyakinsak dizi

A sequence whose terms eventually stay within every band around a single finite number, its limit.

iraksak dizi

A sequence with no finite limit, either because it grows without bound or because it oscillates.

limit of a sequencedizinin limiti

The number the terms close in on as $n$ grows; it need not be a value any term attains.

artan dizi

A sequence with $a_{n+1}\ge a_n$ for every $n$; decreasing is the reverse inequality.

monotonic sequencemonoton dizi

A sequence that is either increasing for all $n$ or decreasing for all $n$.

bounded sequencesinirli dizi

A sequence whose terms all lie between a fixed floor and a fixed ceiling.

geometric sequencegeometrik dizi

A sequence of powers $r^{\,n}$ of a fixed ratio $r$; it converges exactly when $-1<r\le 1$.

recursively defined sequenceozyinelemeli dizi

A sequence given by a starting term and a rule $a_{n+1}=f(a_n)$ that builds each term from the previous one.

Squeeze Theoremsikistirma teoremi

If a sequence is trapped between two others that share a limit, it has that same limit.

Monotonic Sequence Theoremmonoton dizi teoremi

Every sequence that is both monotonic and bounded converges.

factorialfaktoriyel

The product $n!=1\cdot 2\cdots n$; it eventually outgrows any fixed power and any geometric sequence.

What comes next
§02 · Series: adding infinitely many terms

A sequence gave us a list and asked where it lands. Next we add a sequence up, term after term, and ask whether an unending sum can total a finite number. The geometric sequence $r^{\,n}$ you just met becomes the geometric series, the one infinite sum you can evaluate exactly, and the test bed for everything after it.

Sources
  • James Stewart, Calculus (Metric Version), 9th edition, section 11.1 (Sequences) Definitions, the limit laws for sequences, the geometric and standard limits, and the Monotonic Sequence Theorem follow this text, the required book shared with Calculus I.
  • MATH 102 course syllabus: week 1 plan and assessment weights The week 1 line is Sequences and Series 11.1; the published weights are quiz 10% and homework 10%.
  • Standard results of first-year calculus: L'Hopital's Rule and the defining limit of e Carried over from Calculus I and used here as tools; not re-proved.

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