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Week 3Stewart §11.3, 11.4, 11.572 min full read
6 concepts16 worked examples24 exercises3 exam-level4 figures
What are you here for?

03The Integral Test, the Comparison Tests, and Alternating Series

The quiz gives you three sums that all fade to nothing term by term: $\sum \tfrac1{n^2}$, $\sum \tfrac1{n}$, and $\sum \tfrac{(-1)^{n-1}}{n}$. One adds up to a finite number, one runs off to infinity, and one only settles because of its minus signs. Nothing about the shrinking terms alone tells them apart, and that is exactly the gap this week closes.

By the end you can look at a positive series and pick the one test that decides it, read off a in one glance, and put a guaranteed decimal bound on a sum you cannot add up by hand.

In 60 seconds

Four tests decide almost every series in this course: the Integral Test and its p-series shortcut, the two Comparison Tests, and the Alternating Series Test.

Integral Test
$$\sum a_n \text{ and } \int_1^\infty f\,dx \text{ agree}$$

terms are positive and decreasing and $f(x)$ is easy to integrate

p-series
$$\sum \tfrac1{n^p}\ \text{converges}\iff p>1$$

the term is a single power of $n$, or a yardstick for comparison

Comparison / Limit Comparison
$$0\le a_n\le b_n\ \text{ or }\ \tfrac{a_n}{b_n}\to c$$

the term looks like a p-series or geometric series for large $n$

Alternating Series Test
$$b_n\downarrow 0\Rightarrow \sum(-1)^{n-1}b_n\text{ conv.}$$

signs alternate and the sizes decrease to zero

Three most common mistakes

  1. Reading the Integral Test value as the sum: it decides convergence, it is not the total.

  2. Comparing the wrong way, so the inequality you prove says nothing about your series.

  3. Calling an alternating series without checking that the sizes decrease to zero.

Quizzes and homework carry this material; expect to name the test and justify each hypothesis, since a right answer with the wrong reason loses the marks.

How much time do you have?
10 minutes

You walk in able to spot a p-series, run one comparison, and quote the alternating rule.

card, c-p-series, c-comparison, formula card
45 minutes

You can also set up the Integral Test cleanly and choose between direct and limit comparison without guessing.

card, c-integral-test, c-p-series, c-comparison, c-limit-comparison, c-alternating, scaffolding ladder, practice C
full read

Everything above plus the remainder estimates that turn convergence into a guaranteed decimal, and the mixed practice that decides whether you can pick a test unprompted.

all concepts, exam example, practice A, practice B, practice C, practice D
By the end of this section

  1. Apply the Integral Test to a positive decreasing series and state the hypotheses you relied on.

  2. Classify a p-series as convergent or from its exponent alone, and use it as a comparison yardstick.

  3. Prove convergence or divergence with the Direct Comparison Test, choosing an inequality that points the right way.

  4. Use the Limit Comparison Test when a clean inequality is hard, and read the ratio limit correctly.

  5. Test an alternating series for convergence by checking that the sizes decrease to zero.

  6. Estimate a convergent sum and bound the error, using the integral remainder or the alternating remainder.

Syllabus coverage
11.3

The Integral Test and estimates of sums

The test itself and the p-series shortcut it produces sit here; the estimate-of-sums half of 11.3 is the integral remainder in the estimates block.

covered
11.4

The Comparison Tests

Direct comparison first, then the limit comparison test as the tool for when a clean inequality is awkward.

covered
11.5

Alternating series

The Alternating Series Test and its estimation theorem; absolute versus conditional convergence belongs to the next section and is not assumed here.

covered
Recall first
Convergence of a series

$\sum a_n$ converges to $S$ when its $s_N=\sum_{n=1}^N a_n$ have the finite limit $S$. Otherwise it diverges.

Every test in this section is a statement about those partial sums, reached without computing them directly.

The n-th term test for divergence

If $\lim_{n\to\infty} a_n \neq 0$ (or fails to exist), then $\sum a_n$ diverges. If $a_n\to 0$, this test is silent.

It is the first filter and the reason the harder tests exist: the interesting series are exactly the ones whose terms do go to zero.

Geometric series

$\sum_{n=0}^\infty ar^n=\dfrac{a}{1-r}$ when $\lvert r\rvert<1$, and diverges when $\lvert r\rvert\ge 1$.

It is the second yardstick, alongside the p-series, that the comparison tests hold a messy term up against.

Improper integral

$\int_1^\infty f(x)\,dx=\lim_{t\to\infty}\int_1^t f(x)\,dx$; it converges when that limit is finite.

The Integral Test is exactly the bridge from this limit, which you can compute, to the series limit, which you cannot.

Try it yourself first (2 questions)
1§03.0 — does shrinking guarantee a sum●●○○○

Here is the intuition this whole week has to correct. A student argues: the terms of the $\sum \frac1n$ get as small as you like, so the running total cannot keep growing and the series must converge.

Given

  • the series is $\sum_{n=1}^\infty \frac1n$

  • its terms satisfy $\frac1n\to 0$

Find

  1. (a) Is the claim that the series converges true or false?

Hint 1/4

Separate two questions the student has merged: do the terms go to zero, and does the sum stay finite? They are not the same question.

Hint 2/4

The n-th term test only says $a_n\to0$ is necessary. It never promises the sum is finite, so a term going to zero settles nothing on its own.

Hint 3/4

Group the terms: $\frac13+\frac14>\frac12$, and $\frac15+\cdots+\frac18>\frac12$. Each block of this kind adds more than $\frac12$, and there are infinitely many blocks.

Hint 4/4

So the partial sums pass every bound: the harmonic series diverges, and the claim is false.

Show solution
Bound blocks of terms from below
$$\tfrac13+\tfrac14>\tfrac14+\tfrac14=\tfrac12$$

each block of terms is replaced by its smallest member, making the total only smaller

$$\tfrac15+\tfrac16+\tfrac17+\tfrac18>4\cdot\tfrac18=\tfrac12$$

the same trick on the next block of four

$$s_{2^k}>1+\tfrac{k}{2}$$

there are $k$ such blocks up to $2^k$, each worth more than $\tfrac12$

Answer $$\boxed{\text{False: } \sum \tfrac1n \text{ diverges}}$$
Check

The partial sums exceed $1+k/2$, which itself has no finite limit, so no finite ceiling exists.

Terms tending to zero is the entrance fee, not the prize: it lets a series compete, nothing more.

2§03.0 — a geometric warm up●○○○○

Before the new tests, one series you can already total exactly. It appears constantly as a comparison yardstick, so the arithmetic should be automatic.

Given

  • the series is $\sum_{n=1}^{\infty}\left(\tfrac13\right)^{n}$

Find

  1. (a) Does it converge, and if so to what sum?

Hint 1/4

Recognise the shape: a constant ratio between consecutive terms means a geometric series, which you can sum in closed form.

Hint 2/4

For $\sum_{n=1}^\infty ar^{n-1}$ with first term $a$ and ratio $r$, the sum is $\frac{a}{1-r}$ when $\lvert r\rvert<1$.

Hint 3/4

Here the first term (at $n=1$) is $a=\frac13$ and the ratio is $r=\frac13$, so restate: $a=\tfrac13$, $r=\tfrac13$.

Hint 4/4

So the sum is $\dfrac{1/3}{1-1/3}=\dfrac12$.

Show solution
Apply the closed form
$$S=\frac{a}{1-r}=\frac{1/3}{1-1/3}$$

the geometric sum formula, valid because $\lvert r\rvert<1$

$$=\frac{1/3}{2/3}=\frac12$$

clear the compound fraction

Answer $$\boxed{S=\tfrac12}$$
Check

Add the first few terms: $0.333+0.111+0.037+\cdots$ is already $0.481$ and still climbing toward $0.5$.

Every geometric series with $\lvert r\rvert<1$ is a known convergent yardstick for the comparison tests ahead.

Notation
symbolreads asmeanswatch out
$\sum a_n$

the sum of a n

the $a_1+a_2+a_3+\cdots$, shorthand for the limit of its partial sums

the same symbol names the process and its value; $\sum a_n$ 'diverges' means that value does not exist

$s_N$

s sub capital N

the $N$-th partial sum $a_1+\cdots+a_N$, a finite number you can actually add up

lower-case $a_N$ is one term; $s_N$ is the running total of the first $N$

$R_n$

R sub n

the remainder $S-s_n$: everything the tail $a_{n+1}+a_{n+2}+\cdots$ still contributes after $n$ terms

$R_n$ is what you have not yet added, so a small $R_n$ means $s_n$ is already a good estimate

$p$

p

the exponent in a p-series $\sum 1/n^p$; the single number that decides convergence

the cutoff $p>1$ is strict, so $p=1$ (the harmonic series) diverges

Conventions used here
Where the index starts does not decide convergence

Whether a sum begins at $n=1$ or $n=5$ changes the total but never whether it converges: dropping finitely many terms only shifts the sum by a finite amount. So 'for all $n$' in a test may be relaxed to 'for all $n$ beyond some point'.

The Integral Test value is not the sum

When the Integral Test applies, $\int_1^\infty f\,dx$ and $\sum a_n$ share a fate, convergent or divergent, but their numerical values are different. The integral decides; it does not total the series.

Divergence means the partial sums have no finite limit

A positive series either converges to a finite number or diverges to $+\infty$; there is no oscillating middle case. For an alternating series, divergence usually shows up as terms that fail to reach zero.

3.1The Integral Test

Match the terms to a curve you can integrate; a finite area under it means a finite sum.

Where we stand: we can kill a series when its terms miss zero, and total a geometric one exactly. Everything in between — starting with $\sum 1/n^2$ — has no verdict yet.

Solvable with what we have

  • total a geometric series such as $\sum (1/2)^n$ exactly

  • reject $\sum \frac{n}{n+1}$ at a glance, since its terms tend to $1$, not $0$

  • compute any partial sum $s_N$ of $\sum \frac1{n^2}$ by hand

Not solvable yet

  • decide $\sum \frac1{n^2}$, whose terms do tend to $0$

  • decide the harmonic series $\sum \frac1n$, whose terms also tend to $0$

  • say whether the partial sums of $\sum \frac1n$ ever stop climbing

The terms of $\sum \frac1n$ shrink to $0$, so the running total must level off, so the series converges.

Why it fails

The partial sums keep climbing without bound: group them as $1+\tfrac12+\left(\tfrac13+\tfrac14\right)+\left(\tfrac15+\cdots+\tfrac18\right)+\cdots$, and every bracket totals more than $\tfrac12$. Adding more than $\tfrac12$ infinitely often passes every ceiling. Shrinking terms are necessary for convergence but never sufficient; what settles it is how fast the terms shrink, and the area under a matching curve measures exactly that speed.

TheoremTheorem 11.3.1: The Integral Test
Conditions

  • $f$ is continuous, positive, and decreasing on $[1,\infty)$

  • the terms come from that function: $a_n=f(n)$

  • these need hold only for $n\ge N$, since the first few terms never change convergence

$$\boxed{\ \sum_{n=1}^{\infty} a_n \ \text{converges}\iff\ \int_{1}^{\infty} f(x)\,dx \ \text{converges}\ }$$

If the area under the matching curve is finite, the sum is finite; if the area is infinite, the sum runs off to infinity as well.

Looks like this, but is not

$\int_1^\infty \frac{dx}{x^2}=1$, so the sum must be $\sum \frac1{n^2}=1$.

The test compares the sum to the integral; it never equates them. The integral is exactly $1$, but the sum is $\pi^2/6\approx 1.6449$. The Integral Test reports 'finite', not 'how much'. The rectangles overshoot the curve on the left, so the sum sits above the integral.

Deciding the series of 1 over n squared

Use the Integral Test to decide whether $\sum_{n=1}^\infty \frac1{n^2}$ converges.

Given

  • $a_n=\frac1{n^2}$

  • the matching function is $f(x)=\frac1{x^2}$

Find

convergence or divergence

Solution
Check the three hypotheses
$$f(x)=x^{-2}>0\ \text{on}\ [1,\infty)$$

positive, so the rectangles have real height

$$f'(x)=-2x^{-3}<0$$

the derivative is negative, so $f$ is decreasing as required

Integrate over the infinite range
$$\int_1^t x^{-2}\,dx=\Big[-x^{-1}\Big]_1^t=1-\tfrac1t$$

antiderivative of $x^{-2}$ is $-x^{-1}$

$$\lim_{t\to\infty}\left(1-\tfrac1t\right)=1$$

the improper integral converges to a finite value

Read off the verdict
$$\int_1^\infty x^{-2}\,dx=1<\infty\ \Rightarrow\ \sum \tfrac1{n^2}\ \text{converges}$$

finite integral forces a finite sum, by the theorem

Answer $$\boxed{\sum \tfrac1{n^2}\ \text{converges}}$$
Check

Sign and size check: the integral came out positive and finite, and the true sum $\pi^2/6\approx1.645$ is finite and a bit above $1$, exactly as the rectangles predict.

one improper integral, evaluated once.

A finite integral is a certificate of convergence, but its number ($1$ here) is not the sum.

A logarithm hiding a divergence

Decide $\sum_{n=2}^\infty \frac{1}{n\ln n}$ with the Integral Test.

Given

  • $a_n=\frac1{n\ln n}$

  • $f(x)=\frac1{x\ln x}$ on $[2,\infty)$

Find

convergence or divergence

Solution
Confirm the hypotheses on the shifted start
$$x\ln x>0\ \text{for}\ x\ge2$$

both factors are positive past $x=2$, so $f>0$

$$x\ln x\ \text{increases}\ \Rightarrow\ f=\tfrac1{x\ln x}\ \text{decreases}$$

a growing denominator makes the fraction shrink

Integrate by substitution
$$u=\ln x,\quad du=\tfrac1x\,dx$$

the extra $1/x$ in the term is exactly $du$, which is why this substitution is the cheap one

$$\int \frac{dx}{x\ln x}=\int \frac{du}{u}=\ln\lvert u\rvert=\ln(\ln x)$$

the integral collapses to the logarithm of $u$

$$\lim_{t\to\infty}\big[\ln(\ln x)\big]_2^t=\infty$$

the outer logarithm still grows without bound, only very slowly

Read off the verdict
$$\int_2^\infty \frac{dx}{x\ln x}=\infty\ \Rightarrow\ \sum \frac1{n\ln n}\ \text{diverges}$$

an infinite integral forces divergence

Answer $$\boxed{\sum \tfrac1{n\ln n}\ \text{diverges}}$$
Check

Cross-check the direction: the terms $\frac1{n\ln n}$ are larger than a p-series with $p$ just above $1$ would suggest, and $\ln(\ln x)\to\infty$ confirms the slow escape to infinity.

one substitution, one improper integral.

A logarithm in the denominator is the classic near-miss: the terms shrink faster than $1/n$ yet still not fast enough.

Checkpoint
§03.1 — what the integral value means●●○○○

You verify the hypotheses for $\sum 1/n^2$ and compute the matching integral $\int_1^\infty \frac{dx}{x^2}=1$.

Given

  • the Integral Test hypotheses hold

  • $\int_1^\infty x^{-2}\,dx=1$

Find

  1. (a) What may you correctly conclude?

Hint 1/4

Ask what the theorem actually promises: a shared verdict, or a shared number? Those are different claims.

Hint 2/4

The Integral Test says $\sum a_n$ and $\int_1^\infty f$ converge or diverge together; it says nothing about their values matching.

Hint 3/4

Here the integral is finite (it equals $1$), so restate what that licenses: the sum is finite, i.e. it converges.

Hint 4/4

So the sum converges to some finite number, which happens to be $\pi^2/6\approx1.6449$, not $1$.

Show solution
Separate verdict from value
$$\int_1^\infty f<\infty\ \Rightarrow\ \sum a_n\ \text{converges}$$

the theorem transfers finiteness, not the number

$$\sum \tfrac1{n^2}=\tfrac{\pi^2}{6}\approx1.6449\neq1$$

the actual sum differs from the integral

Answer $$\boxed{\text{Converges; sum}\neq1}$$
Check

The rectangles overshoot the curve on the left, so the sum must sit above the integral, and $1.6449>1$ agrees.

⚠ Reporting the integral as the sum

the integral is the number you just computed, so it is tempting to hand it in as the total

wrong$$\sum \tfrac1{n^2}=\int_1^\infty x^{-2}\,dx=1$$
right$$\sum \tfrac1{n^2}=\tfrac{\pi^2}{6}\approx1.6449$$
⚠ Skipping the decreasing check

positivity is obvious, so the decreasing hypothesis gets assumed rather than shown

wrong$$a_n>0\ \Rightarrow\ \text{Integral Test applies}$$
right$$f>0\ \text{and}\ f'<0\ \text{on}\ [N,\infty)$$

3.2The p-series shortcut

Run the Integral Test once on the whole family $\sum 1/n^p$ and you get a one-glance rule you use in every comparison.

The Integral Test is worth running once on the entire family $\sum 1/n^p$, because the answer becomes a lookup we reuse all week.

TheoremTheorem 11.3.2: The p-series test
Conditions

  • the term is a single power: $a_n=\dfrac1{n^p}$

  • $p$ is a real constant

$$\boxed{\ \sum_{n=1}^{\infty}\frac1{n^{p}}\ \text{converges}\iff\ p>1\ }$$

One over n to the p adds to a finite total exactly when the exponent is strictly bigger than one.

Looks like this, but is not

$p=1$ is close enough to the convergent zone, so the harmonic series $\sum 1/n$ should just barely converge.

The cutoff is strict. At $p=1$ the integral is $\ln t\to\infty$, so $\sum 1/n$ diverges. Even $p=1.0001$ converges while $p=1$ does not: the boundary belongs to the divergent side.

Nsum 1/n^2 (p=2)sum 1/n (p=1)sum 1/sqrt(n) (p=1/2)

10

1.5498

2.9290

5.0210

100

1.6350

5.1874

18.5896

1000

1.6439

7.4855

61.8010

10000

1.6448

9.7876

198.5446

The first column is closing in on $\pi^2/6\approx1.6449$. The other two never stop growing, one slowly ($p=1$) and one faster ($p=\tfrac12$), which is exactly the $p>1$ cutoff in numbers.

Two powers, opposite fates

Classify $\sum \frac1{n^{3}}$ and $\sum \frac1{\sqrt{n}}$.

Given

  • first term $\frac1{n^3}$, so $p=3$

  • second term $\frac1{\sqrt n}=n^{-1/2}$, so $p=\tfrac12$

Find

convergence for each

Solution
Read the exponent and compare to 1
$$\sum \tfrac1{n^3}:\ p=3>1\ \Rightarrow\ \text{converges}$$

the exponent clears the cutoff

$$\sum \tfrac1{\sqrt n}=\sum n^{-1/2}:\ p=\tfrac12\le1\ \Rightarrow\ \text{diverges}$$

$\tfrac12$ is on the divergent side of $1$

Answer $$\boxed{\sum \tfrac1{n^3}\ \text{conv.},\quad \sum \tfrac1{\sqrt n}\ \text{div.}}$$
Check

Speed check: cubing the denominator crushes the terms fast, while a square root barely shrinks them, matching convergent and divergent.

The only work in a p-series is writing the term as $n^{-p}$ and reading $p$ off the exponent.

A power just past the cutoff

Decide $\sum \frac{1}{n\sqrt{n}}$ and $\sum \frac{1}{n^{1.001}}$.

Given

  • $\frac1{n\sqrt n}=n^{-3/2}$

  • $\frac1{n^{1.001}}=n^{-1.001}$

Find

convergence for each

Solution
Collapse each term to one power of n
$$\frac1{n\sqrt n}=n^{-1}\cdot n^{-1/2}=n^{-3/2},\ p=\tfrac32>1$$

add the exponents; the result clears $1$, so it converges

$$n^{-1.001},\ p=1.001>1\ \Rightarrow\ \text{converges}$$

just past the cutoff still counts as past it

Answer $$\boxed{\text{both converge}\ (p=\tfrac32,\ p=1.001)}$$
Check

A number check: $\sum n^{-1.001}$ looks like the harmonic series for small $n$, but past the strict cutoff its total is finite, unlike $\sum 1/n$.

Two things you must do first: fold products of powers into one exponent, and never round $1.001$ down to $1$.

Checkpoint
§03.2 — reading the exponent●○○○○

A quiz asks you to classify $\sum 1/n^{0.9}$ in one line, no work shown.

Given

  • the term is $\frac1{n^{0.9}}$, so $p=0.9$

Find

  1. (a) Does the series converge or diverge?

Hint 1/4

This is a pure p-series, so the only thing that matters is where the exponent sits relative to one.

Hint 2/4

The rule: $\sum 1/n^p$ converges when $p>1$ and diverges when $p\le1$.

Hint 3/4

Here $p=0.9$, and restate the comparison: $0.9\le1$.

Hint 4/4

Since $0.9\le1$, the series diverges.

Show solution
Compare the exponent to the cutoff
$$p=0.9\le1\ \Rightarrow\ \text{diverges}$$

the strict cutoff $p>1$ is not met

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

It is termwise larger than the harmonic series ($n^{-0.9}>n^{-1}$), which already diverges, so it must diverge too.

⚠ Making the cutoff non-strict

$p=1$ feels like it should be included with the converging cases next to it

wrong$$\sum \tfrac1{n^p}\ \text{converges for}\ p\ge1$$
right$$\sum \tfrac1{n^p}\ \text{converges for}\ p>1$$
⚠ Leaving the term as a product of powers

$\frac1{n\sqrt n}$ does not look like a p-series until the powers are combined

wrong$$\tfrac1{n\sqrt n}\ \text{is not a p-series}$$
right$$\tfrac1{n\sqrt n}=n^{-3/2},\ p=\tfrac32$$

3.3The Direct Comparison Test

Trap a messy positive term between yourself and a known series, term by term, and inherit its fate.

The p-series and geometric series are known yardsticks. Comparison lets us judge a term we cannot integrate by holding it against one of them, one index at a time.

TheoremTheorem 11.4.1: The Comparison Test
Conditions

  • both series have positive terms: $a_n,b_n>0$

  • the comparison holds eventually: $a_n\le b_n$ for all $n\ge N$

$$\boxed{\ a_n\le b_n:\quad \textstyle\sum b_n\ \text{conv.}\Rightarrow \sum a_n\ \text{conv.};\quad \sum a_n\ \text{div.}\Rightarrow \sum b_n\ \text{div.}\ }$$

A series squeezed under a convergent one also converges; a series propping a divergent one up from below also diverges.

Looks like this, but is not

$\frac1{n^2}\le\frac1n$ and $\sum \frac1n$ diverges, so $\sum \frac1{n^2}$ diverges.

This compares the wrong way. Being below a divergent series says nothing: a short bar under an infinitely tall stack can still add to a finite total. Indeed $\sum 1/n^2$ converges. To prove convergence you must sit under a convergent series; to prove divergence you must sit above a divergent one.

A term tucked under a geometric series

Decide $\sum_{n=1}^\infty \frac1{2^n+n}$ by comparison.

Given

  • $a_n=\frac1{2^n+n}$

  • candidate yardstick $b_n=\frac1{2^n}$

Find

convergence or divergence

Solution
Build the right-pointing inequality
$$2^n+n>2^n\ \Rightarrow\ \frac1{2^n+n}<\frac1{2^n}$$

a bigger denominator makes a smaller fraction; this puts $a_n$ under the yardstick

Name the yardstick's fate and inherit it
$$\sum \frac1{2^n}=\sum\left(\tfrac12\right)^n\ \text{converges}$$

geometric with ratio $\tfrac12<1$

$$a_n<b_n,\ \sum b_n\ \text{conv.}\ \Rightarrow\ \sum a_n\ \text{conv.}$$

convergent ceiling forces convergence

Answer $$\boxed{\sum \tfrac1{2^n+n}\ \text{converges}}$$
Check

The added $n$ only makes each term smaller than the geometric series, so convergence is if anything more secure, not less.

one inequality, one known yardstick.

When a term is a known series plus a positive nuisance in the denominator, drop the nuisance to get the yardstick.

A logarithm that forces divergence

Decide $\sum_{n=1}^\infty \frac{\ln n}{n}$ by comparison.

Given

  • $a_n=\frac{\ln n}{n}$ (positive for $n\ge2$)

  • candidate yardstick $b_n=\frac1n$

Find

convergence or divergence

Solution
Bound the term from below
$$\ln n\ge1\ \text{for}\ n\ge3\ \Rightarrow\ \frac{\ln n}{n}\ge\frac1n$$

past $n=3$ the numerator is at least $1$, putting $a_n$ above the yardstick

Name the yardstick's fate and inherit it
$$\sum \frac1n\ \text{diverges}$$

the harmonic series, $p=1$

$$a_n\ge b_n,\ \sum b_n\ \text{div.}\ \Rightarrow\ \sum a_n\ \text{div.}$$

a divergent floor forces divergence

Answer $$\boxed{\sum \tfrac{\ln n}{n}\ \text{diverges}}$$
Check

Direction check: $\ln n$ grows, so the terms are larger than the already-divergent harmonic terms, and larger-than-divergent must diverge.

one inequality valid past $n=3$, one known yardstick.

For divergence you must sit above a divergent series; the first few terms where the bound fails never matter.

Checkpoint
§03.3 — pointing the inequality the right way●●○○○

You want to prove $\sum \frac1{n^2+n}$ converges by direct comparison, so you need an inequality that points toward a convergent yardstick.

Given

  • $a_n=\frac1{n^2+n}$

Find

  1. (a) Which comparison proves convergence?

Hint 1/4

To prove convergence you must show your term sits under something whose sum you already know is finite.

Hint 2/4

The right yardstick here is a convergent p-series; a divergent one below or above you proves nothing.

Hint 3/4

Since $n^2+n>n^2$, restate the key inequality: $\frac1{n^2+n}<\frac1{n^2}$, and $\sum 1/n^2$ converges.

Hint 4/4

So $\frac1{n^2+n}<\frac1{n^2}$ with a convergent yardstick proves convergence.

Show solution
Match direction to goal
$$n^2+n>n^2\ \Rightarrow\ \frac1{n^2+n}<\frac1{n^2}$$

drop the $+n$ to get a larger, convergent term above

$$\sum \tfrac1{n^2}\ \text{conv.}\ \Rightarrow\ \sum \tfrac1{n^2+n}\ \text{conv.}$$

under a convergent ceiling

Answer $$\boxed{\tfrac1{n^2+n}<\tfrac1{n^2},\ \text{converges}}$$
Check

Limit comparison agrees: the ratio to $1/n^2$ tends to $1$, a positive number, so both share the convergent fate.

⚠ Comparing the wrong way

any true inequality feels like progress, even one that points at the useless yardstick

wrong$$a_n\le \tfrac1n,\ \sum\tfrac1n\ \text{div.}\ \Rightarrow\ \sum a_n\ \text{div.}$$
right$$a_n\le b_n,\ \sum b_n\ \text{conv.}\ \Rightarrow\ \sum a_n\ \text{conv.}$$
⚠ Ignoring that terms must be positive

the test is stated for sizes, and signs get overlooked

wrong$$\text{apply comparison to}\ \sum \tfrac{(-1)^n}{n^2}$$
right$$\text{comparison needs}\ a_n,b_n>0$$

3.4The Limit Comparison Test

When a clean inequality is a fight, compare growth rates instead: a positive finite ratio means the two series share a fate.

Direct comparison needs a clean inequality, and for a term like $\frac{2n^2+3n}{\sqrt{5+n^5}}$ that is real work. Comparing the rate* of shrink sidesteps the algebra.*

TheoremTheorem 11.4.2: The Limit Comparison Test
Conditions

  • both series have positive terms: $a_n,b_n>0$

  • the ratio has a positive finite limit: $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=c,\ 0<c<\infty$

$$\boxed{\ \lim_{n\to\infty}\frac{a_n}{b_n}=c\in(0,\infty):\quad \textstyle\sum a_n\ \text{and}\ \sum b_n\ \text{share a fate}\ }$$

If two positive series shrink at the same rate, so their ratio settles on a positive number, they converge together or diverge together.

Looks like this, but is not

The ratio $a_n/b_n\to0$, so $a_n$ and $b_n$ must still share a fate.

A limit of $0$ (or $\infty$) is outside $(0,\infty)$, and the two-way conclusion fails there. A limit of $0$ only says $a_n$ is eventually much smaller than $b_n$: useful one way (if $\sum b_n$ converges), useless the other. The test's power comes precisely from the ratio being a positive, finite number.

na_nb_n = 1/n^2ratio a_n/b_n

1

2.5000

1.0000

2.5000

5

0.0819

0.0400

2.0480

20

0.005006

0.002500

2.0025

100

0.00020003

0.00010000

2.0003

The ratio homes in on $2$, a positive finite number. So $a_n$ tracks the convergent $1/n^2$ up to a constant, and the series converges with it.

The rate hidden in a root

Decide $\sum_{n=1}^\infty \frac{2n^2+3n}{\sqrt{5+n^5}}$.

Given

  • $a_n=\frac{2n^2+3n}{\sqrt{5+n^5}}$

Find

convergence or divergence

Solution
Find the dominant rate and choose the yardstick
$$a_n\approx\frac{2n^2}{\sqrt{n^5}}=\frac{2n^2}{n^{5/2}}=\frac{2}{n^{1/2}}$$

keep only the highest power top and bottom; that is the rate that survives

$$b_n=\frac{1}{n^{1/2}}$$

strip the constant to get a bare p-series yardstick, $p=\tfrac12$

Take the ratio limit
$$\frac{a_n}{b_n}=\frac{(2n^2+3n)\,n^{1/2}}{\sqrt{5+n^5}}=\frac{2n^{5/2}+3n^{3/2}}{\sqrt{5+n^5}}$$

multiply $a_n$ by $n^{1/2}$; divide top and bottom by $n^{5/2}$ next

$$\longrightarrow \frac{2}{\sqrt{1}}=2\in(0,\infty)$$

leading terms dominate, giving a positive finite limit

Read the shared fate
$$\sum \frac1{n^{1/2}}\ \text{diverges}\ (p=\tfrac12)\ \Rightarrow\ \sum a_n\ \text{diverges}$$

same fate as the yardstick

Answer $$\boxed{\sum \frac{2n^2+3n}{\sqrt{5+n^5}}\ \text{diverges}}$$
Check

Rate check: top grows like $n^2$, bottom like $n^{5/2}$, so $a_n$ decays like $n^{-1/2}$ — a divergent p-series rate.

one ratio limit, no inequality to engineer.

The whole method is: read the dominant power to name $b_n$, then confirm with a ratio limit.

A geometric look-alike

Decide $\sum_{n=1}^\infty \frac1{2^n-1}$.

Given

  • $a_n=\frac1{2^n-1}$

  • yardstick $b_n=\frac1{2^n}$

Find

convergence or divergence

Solution
Take the ratio limit against the geometric series
$$\frac{a_n}{b_n}=\frac{2^n}{2^n-1}=\frac1{1-2^{-n}}$$

divide; the $-1$ becomes a vanishing $2^{-n}$

$$\longrightarrow 1\in(0,\infty)$$

a positive finite limit, so the test applies

Read the shared fate
$$\sum \frac1{2^n}\ \text{converges (geometric)}\ \Rightarrow\ \sum a_n\ \text{converges}$$

same fate as the convergent yardstick

Answer $$\boxed{\sum \tfrac1{2^n-1}\ \text{converges}}$$
Check

Direct comparison confirms it: $2^n-1\ge 2^{n-1}$, so $a_n\le 2/2^n$, still a convergent geometric bound.

one ratio limit against a geometric yardstick.

Subtracting a constant from an exponential does not change its rate, so the geometric fate carries over.

Checkpoint
§03.4 — naming the yardstick●●●○○

You meet $\sum \frac{n+2}{n^3+1}$ and reach for limit comparison. The only decision is which p-series to compare against.

Given

  • $a_n=\frac{n+2}{n^3+1}$

Find

  1. (a) Which yardstick $b_n$ works, and what is the verdict?

Hint 1/4

Find the dominant behaviour of the term for large $n$: keep the highest power on top and bottom.

Hint 2/4

Then take $b_n$ to be the bare power that survives, and check the ratio limit is a positive finite number.

Hint 3/4

Here the top behaves like $n$ and the bottom like $n^3$, so restate: $a_n\approx \frac{n}{n^3}=\frac1{n^2}$, suggesting $b_n=\frac1{n^2}$.

Hint 4/4

The ratio tends to $1$, and $\sum 1/n^2$ converges, so the series converges.

Show solution
Dominant rate, then ratio
$$a_n\approx\frac{n}{n^3}=\frac1{n^2}\ \Rightarrow\ b_n=\frac1{n^2}$$

highest powers only

$$\frac{a_n}{b_n}=\frac{(n+2)n^2}{n^3+1}\to1$$

positive finite limit

$$\sum \tfrac1{n^2}\ \text{conv.}\ \Rightarrow\ \text{converges}$$

shared fate

Answer $$\boxed{b_n=\tfrac1{n^2},\ \text{converges}}$$
Check

Degree bookkeeping: bottom degree minus top degree is $3-1=2>1$, the convergent p-series signature.

⚠ Concluding from a ratio of 0 or infinity

any limit feels like a usable answer, but only $(0,\infty)$ gives the two-way conclusion

wrong$$\tfrac{a_n}{b_n}\to0\ \Rightarrow\ \text{same fate}$$
right$$\text{need}\ 0<\lim\tfrac{a_n}{b_n}<\infty$$
⚠ Picking the wrong dominant power

keeping a lower-order term in the estimate points at the wrong p-series

wrong$$\frac{n+2}{n^3+1}\approx\frac1n$$
right$$\frac{n+2}{n^3+1}\approx\frac1{n^2}$$

3.5The Alternating Series Test

When signs alternate and the sizes shrink to zero, cancellation alone forces convergence, even where the sizes' own series diverges.

Every test so far demanded positive terms. Alternating signs change the story: cancellation can rescue a series whose sizes, added blindly, would diverge.

TheoremTheorem 11.5.1: The Alternating Series Test
Conditions

  • the series alternates: $\sum(-1)^{n-1}b_n$ with $b_n>0$

  • the sizes decrease: $b_{n+1}\le b_n$ eventually

  • the sizes vanish: $\displaystyle\lim_{n\to\infty}b_n=0$

$$\boxed{\ b_n\downarrow 0\ \Rightarrow\ \sum_{n=1}^\infty(-1)^{n-1}b_n\ \text{converges}\ }$$

An alternating series whose sizes decrease steadily to zero always converges.

Looks like this, but is not

$\sum(-1)^{n}\frac{3n}{4n-1}$ alternates, so it converges.

Alternating is not enough; the sizes must reach zero. Here $b_n=\frac{3n}{4n-1}\to\frac34\neq0$, so the terms do not vanish and the n-th term test kills the series outright. Check $b_n\to0$ before anything else.

The alternating harmonic series

Decide $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}$.

Given

  • $b_n=\frac1n>0$

Find

convergence or divergence

Solution
Check the two size conditions
$$b_{n+1}=\tfrac1{n+1}<\tfrac1n=b_n$$

the sizes decrease, condition one

$$\lim_{n\to\infty}\tfrac1n=0$$

the sizes vanish, condition two

Apply the test
$$b_n\downarrow0\ \Rightarrow\ \sum \frac{(-1)^{n-1}}{n}\ \text{converges}$$

both conditions met, so the alternating test fires

Answer $$\boxed{\sum \tfrac{(-1)^{n-1}}{n}\ \text{converges}}$$
Check

The signless version $\sum 1/n$ diverges, so the convergence here is entirely the work of the alternating signs, not the sizes.

two one-line checks.

Alternating signs can save a series whose sizes form a divergent series; the harmonic series is the headline example.

When the sizes refuse to vanish

Decide $\sum_{n=1}^\infty (-1)^{n}\,\frac{3n}{4n-1}$.

Given

  • $b_n=\frac{3n}{4n-1}$

Find

convergence or divergence

Solution
Test the vanishing condition first
$$\lim_{n\to\infty}\frac{3n}{4n-1}=\frac34\neq0$$

divide top and bottom by $n$; the limit is not zero

Fall back to the n-th term test
$$a_n=(-1)^n\tfrac{3n}{4n-1}\ \text{does not tend to}\ 0$$

its size settles near $\tfrac34$, so $a_n$ oscillates without vanishing

$$\Rightarrow\ \sum a_n\ \text{diverges}$$

terms that miss zero force divergence, alternating or not

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

The terms hover near $\pm\tfrac34$ forever, so no cancellation can settle the partial sums; divergence is the only option.

one limit.

Always test $b_n\to0$ first: if it fails, the alternating test is irrelevant and the n-th term test ends it.

Checkpoint
§03.5 — the vanishing gate●●○○○

A quiz shows $\sum (-1)^{n}\,\frac{n}{2n+1}$ and asks for a verdict with reason.

Given

  • $b_n=\frac{n}{2n+1}$

Find

  1. (a) Converge or diverge, and why?

Hint 1/4

Before checking that the sizes decrease, check the cheaper condition: do they even reach zero?

Hint 2/4

The Alternating Series Test needs $b_n\to0$. If that fails, the n-th term test takes over.

Hint 3/4

Divide top and bottom by $n$: $\frac{n}{2n+1}\to\frac12$, so restate: the sizes tend to $\tfrac12$, not $0$.

Hint 4/4

Since the sizes miss zero, the terms do not vanish and the series diverges.

Show solution
Check the limit of the sizes
$$\frac{n}{2n+1}\to\frac12\neq0$$

the sizes settle at $\tfrac12$

$$\Rightarrow\ a_n\not\to0\ \Rightarrow\ \text{diverges}$$

n-th term test

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

The size limit $\tfrac12$ is positive, so the terms oscillate near $\pm\tfrac12$ and cannot settle.

⚠ Assuming alternating means convergent

the sign pattern looks like the whole hypothesis

wrong$$\text{alternating}\ \Rightarrow\ \text{converges}$$
right$$\text{need}\ b_n\downarrow\ \text{and}\ b_n\to0$$
⚠ Checking decrease but not vanishing

the decreasing condition is the fussier one, so it grabs the attention

wrong$$b_{n+1}\le b_n\ \Rightarrow\ \text{converges}$$
right$$b_{n+1}\le b_n\ \text{and}\ b_n\to0$$

3.6Estimating a sum and bounding the error

Convergence promises a number exists; the remainder estimates hand you that number to a guaranteed number of decimals.

A convergence verdict says a total exists. What an exam or a computer actually needs is that total to, say, three decimals, with a guarantee on the error.

NoteTwo remainder bounds
Conditions

  • integral remainder: the Integral Test hypotheses hold for $f$

  • alternating remainder: the Alternating Series Test hypotheses hold

$$\boxed{\ \int_{n+1}^{\infty}\! f\,dx\ \le\ R_n\ \le\ \int_{n}^{\infty}\! f\,dx\ }\qquad\boxed{\ \lvert R_n\rvert\le b_{n+1}\ }$$

For a positive decreasing series the leftover sits between the two tail integrals; for an alternating series the leftover is no bigger than the first term you dropped.

Looks like this, but is not

$\lvert R_n\rvert\le b_{n+1}$ is a handy bound, so use it on $\sum \frac1{n^2}$ after $n$ terms.

That bound is for alternating series only. $\sum 1/n^2$ has all-positive terms; its remainder is bounded by the tail integral, not by the next term. Using the wrong remainder rule gives a bound that is not valid.

Pinning down the sum of 1 over n squared

Use $s_{10}$ and the integral remainder to bracket $S=\sum_{n=1}^\infty \frac1{n^2}$.

Given

  • $s_{10}=\sum_{n=1}^{10}\frac1{n^2}=1.549768$

  • $f(x)=\frac1{x^2}$, so $\int_n^\infty f=\frac1n$

Find

an interval containing $S$

Solution
Bound the remainder both ways
$$\int_{11}^\infty \frac{dx}{x^2}=\frac1{11}\le R_{10}\le \int_{10}^\infty \frac{dx}{x^2}=\frac1{10}$$

the two tail integrals, from $\int_n^\infty x^{-2}=1/n$

$$0.0909\le R_{10}\le0.1000$$

as decimals

Add the known partial sum
$$s_{10}+\tfrac1{11}\le S\le s_{10}+\tfrac1{10}$$

since $S=s_{10}+R_{10}$

$$1.6407\le S\le1.6498$$

arithmetic

Answer $$\boxed{1.6407\le S\le1.6498}$$
Check

The true value $\pi^2/6=1.64493$ lies inside the bracket, confirming the bound is honest.

one partial sum plus two one-line integrals.

Averaging the two ends, $S\approx1.645$, is already good to three decimals from just ten terms and one integral.

How many terms for three-decimal accuracy

For $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}$, how many terms guarantee an error below $0.001$?

Given

  • alternating with $b_n=\frac1{n^3}$

  • target error $<0.001$

Find

the smallest number of terms $n$

Solution
Set the alternating remainder below the target
$$\lvert R_n\rvert\le b_{n+1}=\frac1{(n+1)^3}$$

the estimation theorem, since the test's hypotheses hold

$$\frac1{(n+1)^3}<0.001=\frac1{1000}\ \Rightarrow\ (n+1)^3>1000$$

clear the fraction

Solve for n
$$(n+1)^3>1000\ \Rightarrow\ n+1>10\ \Rightarrow\ n\ge10$$

cube root; $n+1=10$ gives exactly $0.001$, not less, so go one higher

Answer $$\boxed{n=10\ \text{terms}}$$
Check

Check the boundary: $n=9$ gives $b_{10}=1/1000=0.001$, not strictly below; $n=10$ gives $b_{11}=1/1331\approx0.00075<0.001$.

one inequality, one cube root.

The alternating bound is the cheapest error estimate in the course: the first dropped term is the bound.

Checkpoint
§03.6 — the first dropped term●●○○○

You add the first four terms of $\sum \frac{(-1)^{n-1}}{n}$ and stop. You want a guaranteed bound on how far $s_4$ is from the true sum.

Given

  • alternating with $b_n=\frac1n$

  • you have summed $n=4$ terms

Find

  1. (a) What is the guaranteed error bound $\lvert R_4\rvert$?

Hint 1/4

The error after stopping an alternating series is controlled by one specific term, not by the whole tail.

Hint 2/4

The estimation theorem says $\lvert R_n\rvert\le b_{n+1}$: the size of the first term you did not add.

Hint 3/4

You summed through $n=4$, so the first dropped term is $b_5=\frac15$; restate: $\lvert R_4\rvert\le b_5$.

Hint 4/4

So the bound is $\lvert R_4\rvert\le\frac15=0.2$.

Show solution
Apply the estimation theorem
$$\lvert R_4\rvert\le b_{5}=\frac15$$

the bound is the first omitted term

$$=0.2$$

as a decimal

Answer $$\boxed{\lvert R_4\rvert\le\tfrac15}$$
Check

Direct check: $s_4=0.5833$ and $S=\ln2=0.6931$, a gap of $0.11<0.2$, inside the bound.

⚠ Using the alternating bound on a positive series

$b_{n+1}$ is the easiest bound to write, so it gets used everywhere

wrong$$\sum\tfrac1{n^2}:\ \lvert R_n\rvert\le \tfrac1{(n+1)^2}$$
right$$\sum\tfrac1{n^2}:\ R_n\le\int_n^\infty x^{-2}dx=\tfrac1n$$
⚠ Bounding by the last term kept

off-by-one: the term you just added feels like the relevant one

wrong$$\lvert R_n\rvert\le b_n$$
right$$\lvert R_n\rvert\le b_{n+1}$$
Choosing a convergence test

you are handed a series and must decide which of this section's tests to reach for

  1. Check the terms reach zero

    If $a_n\not\to0$, stop: the series diverges by the n-th term test. Only bother with the rest when $a_n\to0$.

  2. Look for a single power or an exponential

    A bare $1/n^p$ is a p-series; a bare $r^n$ is geometric. Read the verdict directly.

  3. For a rational or root-of-polynomial term, compare

    Keep the highest power top and bottom to guess the rate, pick that p-series as $b_n$, then confirm with limit comparison.

  4. For alternating signs, use the alternating test

    Check the sizes $b_n$ decrease to $0$. If the sizes fail to reach $0$, it is the n-th term test again.

  5. For a term you can integrate, use the Integral Test

    When $f(x)$ is easy to integrate (logs, exponentials, simple powers), the Integral Test both decides and estimates.

Where it goes wrong

  • jumping to a test before checking $a_n\to0$

  • comparing the wrong way with direct comparison

Running a limit comparison cleanly

the term is a ratio or root of polynomials that looks like a p-series but is not exactly one

  1. Estimate the rate

    Replace top and bottom by their highest powers to see how the term decays, e.g. $\frac{2n^2+3n}{\sqrt{5+n^5}}\approx \frac{2}{n^{1/2}}$.

  2. Strip to a bare p-series

    Drop the constant to name $b_n=1/n^{p}$ with the same rate.

  3. Take the ratio limit

    Compute $\lim a_n/b_n$. A positive finite value licenses the test; $0$ or $\infty$ means you chose the wrong $b_n$.

  4. Quote the yardstick's fate

    State whether $\sum b_n$ converges (as a p-series) and conclude the same for $\sum a_n$.

Where it goes wrong

  • picking $b_n$ with the wrong power so the ratio is $0$ or $\infty$

  • forgetting to say the p-series verdict at the end

The harmonic series diverges

Classify $\sum \frac1n$.

Given

  • $a_n=\frac1n$, and $a_n\to0$

Find

convergence or divergence

Solution
Read it as a p-series
$$\sum \tfrac1n:\ p=1$$

a single power of $n$, exponent $1$

$$p=1\le1\ \Rightarrow\ \text{diverges}$$

the strict cutoff $p>1$ is missed at the boundary

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

The Integral Test agrees: $\int_1^\infty dx/x=\ln t\to\infty$.

The p equals 2 series converges

Classify $\sum \frac1{n^2}$.

Given

  • $a_n=\frac1{n^2}$, and $a_n\to0$

Find

convergence or divergence

Solution
Read it as a p-series
$$\sum \tfrac1{n^2}:\ p=2$$

exponent $2$

$$p=2>1\ \Rightarrow\ \text{converges}$$

clear of the cutoff

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

The Integral Test agrees: $\int_1^\infty dx/x^2=1<\infty$.

Both have positive terms tending to zero and both look like $1/n^{\text{power}}$, yet one diverges and one converges.

How to tell them apart

The only thing separating them is the exponent against the strict cutoff $p>1$: $p=1$ diverges, $p=2$ converges. Shrinking terms are not the decider; the rate is.

Scaffolding comes off
The common skeleton

  1. Estimate the rate: keep the highest power of $n$ on top and bottom to see how the term decays.

  2. Name the yardstick: take $b_n=1/n^p$ (or a geometric $r^n$) with that same rate.

  3. Pick the tool: use limit comparison (a positive finite ratio) unless a clean inequality makes direct comparison faster.

  4. State the verdict: quote whether $\sum b_n$ converges and carry the shared fate to $\sum a_n$.

1 · fully worked

Full worked comparison

Decide $\sum_{n=1}^\infty \frac{n^2+1}{n^4+3}$.

Given

  • $a_n=\frac{n^2+1}{n^4+3}$

Find

convergence or divergence

Solution
Rate and yardstick
$$a_n\approx\frac{n^2}{n^4}=\frac1{n^2}$$

highest powers only

$$b_n=\frac1{n^2}\ (p=2)$$

the bare p-series with that rate

Ratio and verdict
$$\frac{a_n}{b_n}=\frac{(n^2+1)n^2}{n^4+3}\to1\in(0,\infty)$$

positive finite limit, so limit comparison applies

$$\sum \tfrac1{n^2}\ \text{conv.}\ \Rightarrow\ \sum a_n\ \text{conv.}$$

shared fate

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

Degree gap bottom minus top is $4-2=2>1$, the convergent signature.

2 · you write the reasoning

Same skeleton, and the steps are written out. Supply the reason for each before opening it: decide $\sum_{n=1}^\infty \frac{2n+1}{n^2+n}$.

  1. $a_n\approx\frac{2n}{n^2}=\frac2n$.

    reasoning

    Keep only the leading powers: top like $2n$, bottom like $n^2$, so the term decays like $2/n$.

  2. Take $b_n=\frac1n$.

    reasoning

    Strip the constant $2$ to name the bare p-series yardstick of the same rate.

  3. $\frac{a_n}{b_n}=\frac{(2n+1)n}{n^2+n}\to2$.

    reasoning

    A positive finite ratio ($2$) is exactly what limit comparison needs to link the two fates.

  4. $\sum\frac1n$ diverges, so the series diverges.

    reasoning

    The yardstick is the harmonic series ($p=1$), which diverges, so the series diverges with it.

3 · find the buried error

Harder term, and a solution written by someone else. Exactly two of the four steps contain an error. Find both before reading on: decide $\sum_{n=1}^\infty \frac{2n^2+5}{n^3+1}$.

  1. Step 1. For large $n$, $a_n\approx\frac{2n^2}{n^3}=\frac{2}{n^2}$.

  2. Step 2. Take $b_n=\frac1{n^2}$.

  3. Step 3. $\frac{a_n}{b_n}=\frac{(2n^2+5)n^2}{n^3+1}\to2$.

  4. Step 4. Since $\sum\frac1{n^2}$ converges, the series converges.

the two buried errors (2)
⚠ step 1

$\frac{2n^2}{n^3}=\frac{2}{n}$, not $\frac{2}{n^2}$.

subtracting the exponents wrongly ($n^2/n^3=n^{-1}$) is the most common slip in naming the rate.

right

The correct rate is $\frac2n$, so the right yardstick is $b_n=\frac1n$.

⚠ step 3

With $b_n=1/n^2$ the ratio is $\frac{(2n^2+5)n^2}{n^3+1}\to\infty$, not $2$.

writing down a tidy constant without actually dividing top and bottom by the leading power.

right

Against the correct $b_n=1/n$ the ratio is $\frac{(2n^2+5)n}{n^3+1}\to2$, a positive finite number.

4 · the bare problem
§03.4 — a geometric look-alike, unaided●●●○○

No scaffolding this time. Decide the series and name the yardstick you used.

Given

  • $a_n=\frac{3^n}{4^n+1}$

Find

  1. (a) Converge or diverge, with a yardstick?

Hint 1/4

Exponentials dominate polynomials, so look at how the powers of 3 and 4 compare, not the $+1$.

Hint 2/4

A term behaving like $(3/4)^n$ suggests a geometric yardstick $b_n=(3/4)^n$ with ratio below $1$.

Hint 3/4

Divide: $\frac{3^n}{4^n+1}\approx\frac{3^n}{4^n}=(3/4)^n$, so restate $b_n=(3/4)^n$ and take the ratio.

Hint 4/4

The ratio $\frac{a_n}{b_n}=\frac{4^n}{4^n+1}\to1$, and $\sum(3/4)^n$ converges, so the series converges.

Show solution
Geometric yardstick and ratio
$$b_n=(3/4)^n$$

the dominant exponential rate

$$\frac{a_n}{b_n}=\frac{4^n}{4^n+1}\to1$$

positive finite limit

$$\sum(3/4)^n\ \text{conv.}\Rightarrow\ \text{converges}$$

geometric with ratio $<1$

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

Direct comparison agrees: $\frac{3^n}{4^n+1}<(3/4)^n$, a convergent geometric bound.

Full exam-style question

Name the test for each of three seriesexam format

Classify each series as convergent or divergent, and name the test you used: (a) $\sum \frac{n}{n^2+1}$, (b) $\sum_{n=2}^\infty \frac1{n(\ln n)^2}$, (c) $\sum \frac{(-1)^{n-1}\,n}{n^2+1}$.

Given

  • three series, (a) positive, (b) positive, (c) alternating

Find

a verdict and a named test for each part

Solution
(a) Limit comparison with the harmonic series
$$\frac{n}{n^2+1}\approx\frac{n}{n^2}=\frac1n\ \Rightarrow\ b_n=\frac1n$$

dominant rate is $1/n$

$$\frac{a_n}{b_n}=\frac{n^2}{n^2+1}\to1$$

positive finite limit

$$\sum\tfrac1n\ \text{div.}\ \Rightarrow\ \text{(a) diverges}$$

shared fate with the harmonic series

(b) Integral Test with a substitution
$$u=\ln x,\ du=\tfrac{dx}{x}\ \Rightarrow\ \int\frac{dx}{x(\ln x)^2}=\int\frac{du}{u^2}=-\frac1{\ln x}$$

the $1/x$ supplies $du$

$$\int_2^\infty\frac{dx}{x(\ln x)^2}=\Big[-\tfrac1{\ln x}\Big]_2^\infty=\frac1{\ln 2}<\infty$$

a finite improper integral

$$\Rightarrow\ \text{(b) converges}$$

finite integral, so finite sum

(c) Alternating Series Test
$$b_n=\frac{n}{n^2+1}\to0$$

the sizes vanish

$$\frac{d}{dx}\frac{x}{x^2+1}=\frac{1-x^2}{(x^2+1)^2}<0\ (x>1)$$

the sizes decrease past $n=1$

$$\Rightarrow\ \text{(c) converges}$$

both alternating conditions hold

Answer $$\boxed{\text{(a) diverges,\ (b) converges,\ (c) converges}}$$
Check

Consistency check: (a) and (c) share the same sizes $n/(n^2+1)\approx1/n$, yet (a) diverges and (c) converges — the alternating signs in (c) are doing the work, exactly the theme of this section.

one limit comparison, one integral with a substitution, one alternating check.

On an exam, naming the test and verifying its hypotheses earns the marks; a bare 'converges' with no test named usually does not.

Practice

A · concept 3 questions
1§03.1 — does vanishing settle it●○○○○

A classmate writes: the terms of $\sum \frac1{\sqrt n}$ go to zero, so the series converges. Judge the reasoning, not just the answer.

Given

  • $a_n=\frac1{\sqrt n}\to0$

Find

  1. (a) True or false: the series converges?

Hint 1/4

Terms going to zero is only the entry condition; it never decides convergence by itself.

Hint 2/4

Read it as a p-series: $\frac1{\sqrt n}=n^{-1/2}$, and the rule is convergence iff $p>1$.

Hint 3/4

Here $p=\frac12$, so restate: $p=\frac12\le1$.

Hint 4/4

With $p\le1$ the series diverges, so the claim is false.

Show solution

The term is a single power of $n$, so we read the exponent instead of testing terms one by one; a comparison here would only re-derive the p-series rule.

Write the term as a power of n
$$\frac1{\sqrt n}=n^{-1/2},\ p=\tfrac12$$

a root is a fractional power, so naming $p$ is all the p-series rule needs

Compare the exponent to the cutoff
$$p=\tfrac12\le1\ \Rightarrow\ \text{diverges}$$

the cutoff $p>1$ is strict and $\tfrac12$ falls on the divergent side, so the claim is false

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

It is termwise larger than the harmonic series, which already diverges.

Terms going to zero is the entry ticket, never the verdict; for a bare power the exponent alone decides.

2§03.1 — the one thing convergence guarantees●○○○○

Another claim: if $\sum a_n$ converges, then $a_n\to0$. Decide whether this direction holds.

Given

  • $\sum a_n$ is given to converge

Find

  1. (a) True or false?

Hint 1/4

This is the reverse of the n-th term test; ask whether a finite total can tolerate terms that do not shrink.

Hint 2/4

If the partial sums approach a limit $S$, then $a_n=s_n-s_{n-1}\to S-S=0$.

Hint 3/4

So convergence forces $a_n\to0$; restate: the difference of two things heading to $S$ heads to $0$.

Hint 4/4

Hence the statement is true.

Show solution

We argue straight from the definition of convergence rather than from any test, because the claim is about what convergence itself forces on the terms.

Write each term as a difference of partial sums
$$a_n=s_n-s_{n-1}$$

every term is the jump between consecutive running totals, by the definition of $s_n$

Take the limit of both partial sums
$$s_n\to S,\ s_{n-1}\to S\ \Rightarrow\ a_n\to S-S=0$$

both totals chase the same limit $S$, so their difference is squeezed to zero and the claim holds

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

This is exactly why the n-th term test is valid as a divergence test.

This is a one-way street: convergence forces $a_n\to0$, but $a_n\to0$ never forces convergence, which is why the n-th term test can only reject.

3§03.2 — just past the cutoff●●○○○

Speed round: classify $\sum \frac1{n^{1.01}}$, which looks deceptively like the harmonic series for small $n$.

Given

  • $a_n=n^{-1.01}$

Find

  1. (a) Converge or diverge?

Hint 1/4

Ignore how it looks for small $n$; convergence is about the tail and the exponent decides it.

Hint 2/4

The p-series rule: convergence iff $p>1$, no matter how close $p$ is to $1$.

Hint 3/4

Here $p=1.01$, so restate: $1.01>1$.

Hint 4/4

Since $p>1$, the series converges.

Show solution

A single power of $n$ is settled by its exponent alone, so we skip any comparison and read $p$ off directly.

Read the exponent
$$a_n=n^{-1.01},\ p=1.01$$

the term is already one power of $n$, so $p$ is visible without rewriting

Compare to the strict cutoff
$$p=1.01>1\ \Rightarrow\ \text{converges}$$

$1.01$ clears $1$, so the tail is summable and the small-$n$ resemblance to $1/n$ is irrelevant

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

Its tail integral $\int_1^\infty x^{-1.01}dx=1/0.01=100$ is finite.

Never round the exponent: $1.01$ and $1$ sit on opposite sides of the cutoff even though they look almost equal.

B · computation 6 questions
1§03.1 — a clean Integral Test●●○○○

Set up and run the Integral Test on a term built for it.

Given

  • $\sum_{n=1}^\infty n e^{-n^2}$

  • $f(x)=x e^{-x^2}$, positive and eventually decreasing

Find

  1. (a) Converge or diverge?

Hint 1/4

The extra factor of $x$ in front is a hint that a substitution will integrate this cleanly.

Hint 2/4

Use the Integral Test: evaluate $\int_1^\infty x e^{-x^2}\,dx$ and read off its finiteness.

Hint 3/4

Let $u=x^2$, $du=2x\,dx$, so $\int x e^{-x^2}dx=\frac12\int e^{-u}du$; restate the integral in $u$.

Hint 4/4

The integral is $\frac1{2e}$, finite, so the series converges.

Show solution

The factor $x$ out front is exactly the derivative of $x^2$, so $u=x^2$ clears the integral in one move; without that matching factor the Integral Test would be far more work.

Substitute to match the front factor
$$u=x^2,\ du=2x\,dx$$

$x\,dx=\tfrac12\,du$, so the loose $x$ is absorbed exactly, which is why this substitution is the cheap route

Evaluate the improper integral
$$\int_1^\infty xe^{-x^2}\,dx=\tfrac12\big[-e^{-u}\big]_1^\infty=\tfrac1{2e}$$

the exponential decays to $0$ at the top, leaving a finite value

Read off the verdict
$$\tfrac1{2e}<\infty\ \Rightarrow\ \text{converges}$$

a finite matching integral forces a finite sum, by the Integral Test

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

The value $1/(2e)\approx0.184$ is finite and positive, as an area should be.

When a stray factor equals the derivative of an inner expression, the Integral Test is almost free; look for that match first.

2§03.2 — read the exponent●○○○○

A pure p-series, hidden behind a cube root.

Given

  • $\sum_{n=1}^\infty \frac1{\sqrt[3]{n^2}}=\sum n^{-2/3}$

Find

  1. (a) Converge or diverge?

Hint 1/4

Rewrite the root as a power of $n$ so the exponent is visible.

Hint 2/4

Then apply the p-series rule: convergence iff $p>1$.

Hint 3/4

$\sqrt[3]{n^2}=n^{2/3}$, so restate: $p=\frac23$.

Hint 4/4

Since $p=\frac23\le1$, it diverges.

Show solution

A root is just a fractional power, so we rewrite it as $n^{-p}$ and use the p-series rule instead of hunting for a comparison.

Rewrite the root as a power
$$\frac1{\sqrt[3]{n^2}}=n^{-2/3},\ p=\tfrac23$$

a cube root of $n^2$ is $n^{2/3}$, so $p=\tfrac23$ is now readable

Compare to the cutoff
$$p=\tfrac23\le1\ \Rightarrow\ \text{diverges}$$

$\tfrac23$ is below the strict cutoff, so the series diverges like a shallow p-series

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

Termwise larger than the harmonic series, which diverges.

Roots hide the exponent; convert to $n^{-p}$ first and the verdict is immediate.

3§03.3 — a direct comparison●●○○○

Bound the term above by a convergent yardstick.

Given

  • $\sum_{n=1}^\infty \frac1{n^3+2n}$

Find

  1. (a) Converge or diverge, by direct comparison?

Hint 1/4

Drop the smaller piece of the denominator to make a cleaner, larger term above yours.

Hint 2/4

Direct comparison for convergence: find $b_n\ge a_n$ with $\sum b_n$ convergent.

Hint 3/4

Since $n^3+2n>n^3$, restate: $\frac1{n^3+2n}<\frac1{n^3}$, and $\sum 1/n^3$ converges ($p=3$).

Hint 4/4

Under a convergent yardstick, the series converges.

Show solution

The denominator is a p-series denominator plus a positive nuisance, so a one-line direct comparison beats limit comparison; dropping the nuisance gives the yardstick at once.

Drop the nuisance to build the inequality
$$n^3+2n>n^3\ \Rightarrow\ \frac1{n^3+2n}<\frac1{n^3}$$

a larger denominator makes a smaller fraction, placing the term under the yardstick

Inherit the yardstick's fate
$$\sum \tfrac1{n^3}\ (p=3)\ \text{conv.}\ \Rightarrow\ \text{converges}$$

sitting below a convergent series caps the partial sums, so convergence transfers

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

Limit comparison with $1/n^3$ gives ratio $\to1$, agreeing.

Known series plus a positive term in the denominator: drop the extra term to expose the yardstick and compare directly.

4§03.4 — a limit comparison●●○○○

The polynomials fight a clean inequality, so compare rates instead.

Given

  • $\sum_{n=1}^\infty \frac{3n-2}{n^3+4}$

Find

  1. (a) Converge or diverge, by limit comparison?

Hint 1/4

Read the dominant rate from the highest powers on top and bottom.

Hint 2/4

Take $b_n$ to be that bare power and check the ratio limit lands in $(0,\infty)$.

Hint 3/4

Top like $3n$, bottom like $n^3$, so $a_n\approx\frac3{n^2}$; restate $b_n=\frac1{n^2}$.

Hint 4/4

The ratio tends to $3$, and $\sum 1/n^2$ converges, so the series converges.

Show solution

Top and bottom are polynomials with no clean inequality, so we compare rates with limit comparison rather than forcing a fragile direct bound.

Name the dominant rate
$$a_n\approx\frac{3n}{n^3}=\frac3{n^2}\ \Rightarrow\ b_n=\frac1{n^2}$$

keep only the leading powers; the constant $3$ is dropped because limit comparison ignores positive constant factors

Take the ratio and read it
$$\frac{a_n}{b_n}=\frac{(3n-2)n^2}{n^3+4}\to3$$

a positive finite limit is exactly the hypothesis limit comparison needs to link the two fates

Carry the shared fate
$$\sum \tfrac1{n^2}\ \text{conv.}\ \Rightarrow\ \text{converges}$$

with a finite positive ratio, the series and its yardstick converge together

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

Degree gap $3-1=2>1$, the convergent signature.

For a ratio of polynomials, subtract degrees: bottom minus top above $1$ means convergence, and the leading coefficients only set the ratio, not the verdict.

5§03.5 — an alternating test●●○○○

Signs alternate; check the two size conditions.

Given

  • $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n+1}$

Find

  1. (a) Converge or diverge, by the Alternating Series Test?

Hint 1/4

For an alternating series, isolate the sizes $b_n$ and test two things: do they shrink, and do they reach zero?

Hint 2/4

The test needs $b_{n+1}\le b_n$ and $b_n\to0$.

Hint 3/4

Here $b_n=\frac1{2n+1}$; restate: it decreases and $\frac1{2n+1}\to0$.

Hint 4/4

Both hold, so the series converges.

Show solution

The signs alternate, so we check the two Leibniz conditions rather than any positive-series test, which would not apply to a signed series.

Check the sizes decrease
$$b_{n+1}=\tfrac1{2n+3}<\tfrac1{2n+1}=b_n$$

the denominator grows, so each size is strictly smaller than the last

Check the sizes vanish
$$\lim_{n\to\infty}\tfrac1{2n+1}=0$$

the denominator runs to infinity, so the sizes reach zero

Fire the test
$$b_n\downarrow0\ \Rightarrow\ \text{converges}$$

both Leibniz conditions hold, so the alternating series converges

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

This is the Leibniz series for $\tfrac{\pi}{4}$, known to converge.

For an alternating series, never test the signless term for convergence; only that its size decreases to zero.

6§03.4 — a hidden simplification●●●○○

Before choosing a test, simplify: the term collapses to a familiar one.

Given

  • $\sum_{n=1}^\infty \frac{n+1}{n^3+n^2}$

Find

  1. (a) Converge or diverge?

Hint 1/4

Factor the denominator before comparing; a common factor may cancel the numerator entirely.

Hint 2/4

If the term simplifies to a p-series, read the verdict straight off.

Hint 3/4

$n^3+n^2=n^2(n+1)$, so restate: $\frac{n+1}{n^2(n+1)}=\frac1{n^2}$.

Hint 4/4

So the series is exactly $\sum 1/n^2$, which converges.

Show solution

Before naming any test we factor and cancel: the term collapses to an exact p-series, so a comparison or ratio would be wasted effort.

Factor the denominator
$$n^3+n^2=n^2(n+1)$$

pulling out $n^2$ exposes the shared $(n+1)$ that will cancel

Cancel the common factor
$$\frac{n+1}{n^2(n+1)}=\frac1{n^2}$$

the $(n+1)$ on top kills the one below, leaving a bare power

Classify the survivor
$$p=2>1\ \Rightarrow\ \text{converges}$$

the term is identically $1/n^2$, a $p=2$ series past the cutoff

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

No estimation needed; the term is identically $1/n^2$.

Always simplify before choosing a test; a term that cancels to $1/n^2$ needs no comparison at all.

C · exam level 3 questions
1§03.4 — a root in the denominator●●●○○

An exam favourite: a term that decays just slowly enough to diverge.

Given

  • $\sum_{n=1}^\infty \frac1{\sqrt{n^2+1}}$

Find

  1. (a) Converge or diverge, and by what comparison?

Hint 1/4

Estimate the size for large $n$: the $+1$ under the root is negligible next to $n^2$.

Hint 2/4

Use limit comparison with the p-series that matches that rate.

Hint 3/4

$\sqrt{n^2+1}\approx n$, so $a_n\approx\frac1n$; restate $b_n=\frac1n$.

Hint 4/4

The ratio tends to $1$ and $\sum 1/n$ diverges, so the series diverges.

Show solution

The root blocks a clean inequality against $1/n$, so limit comparison reads the rate directly instead of forcing a bound that barely fails.

Name the dominant rate
$$\sqrt{n^2+1}\approx n\ \Rightarrow\ b_n=\frac1n$$

for large $n$ the $+1$ is negligible under the root, so the term decays like $1/n$

Take the ratio
$$\frac{a_n}{b_n}=\frac{n}{\sqrt{n^2+1}}\to1$$

a positive finite ratio lets limit comparison transfer the verdict

Carry the divergent fate
$$\sum \tfrac1n\ \text{div.}\ \Rightarrow\ \text{diverges}$$

the harmonic yardstick diverges, so the series diverges with it

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

For all $n$, $\sqrt{n^2+1}<n+1$, so $a_n>\frac1{n+1}$, above a divergent tail.

A square root of $n^2$ plus a constant decays like $1/n$; that borderline rate diverges, so read the rate, do not trust the shrinking look.

2§03.5 — alternating over a root●●●○○

The sizes here form a divergent series on their own, so the signs must be doing the work.

Given

  • $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{\sqrt n}$

Find

  1. (a) Converge or diverge?

Hint 1/4

Note first that the sizes $1/\sqrt n$ form a divergent p-series, so any convergence is due to the alternation.

Hint 2/4

Apply the Alternating Series Test to the sizes $b_n=\frac1{\sqrt n}$: do they decrease to zero?

Hint 3/4

$\frac1{\sqrt{n+1}}<\frac1{\sqrt n}$ and $\frac1{\sqrt n}\to0$; restate: sizes decrease to $0$.

Hint 4/4

Both conditions hold, so the alternating series converges.

Show solution

The signless sizes form a divergent p-series, so no positive-series test can help; only the Alternating Series Test, which reads the signed structure, applies.

Check the sizes decrease
$$b_{n+1}=\tfrac1{\sqrt{n+1}}<\tfrac1{\sqrt n}=b_n$$

a growing root in the denominator shrinks each size below the last

Check the sizes vanish
$$b_n=\tfrac1{\sqrt n}\to0$$

the root runs to infinity, so the sizes reach zero

Fire the test
$$b_n\downarrow0\ \Rightarrow\ \text{converges}$$

both Leibniz conditions hold, so the alternation secures convergence even though the sizes' own series diverges

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

The signless $\sum 1/\sqrt n$ ($p=\tfrac12$) diverges, so the signs alone secure convergence.

A divergent signless series can still converge with signs; the alternating test cares only about decrease to zero, not summability of the sizes.

3§03.6 — how many terms for two decimals●●●○○

You will estimate an alternating sum and need a guaranteed accuracy.

Given

  • $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}$

  • target error $<0.01$

Find

  1. (a) How many terms guarantee an error below $0.01$?

Hint 1/4

The error after stopping an alternating series is at most the first term you drop.

Hint 2/4

So require $b_{n+1}=\frac1{(n+1)^2}<0.01$ and solve for $n$.

Hint 3/4

$\frac1{(n+1)^2}<\frac1{100}$ means $(n+1)^2>100$; restate: $n+1>10$.

Hint 4/4

So $n\ge10$: ten terms suffice, with error at most $1/121\approx0.0083$.

Show solution

For an alternating series the remainder is bounded by the first dropped term, so we solve one inequality rather than summing tails.

Bound the error by the first dropped term
$$|R_n|\le b_{n+1}=\frac1{(n+1)^2}<\frac1{100}$$

the alternating remainder never exceeds the next size, so making that size small enough guarantees the accuracy

Solve for n
$$(n+1)^2>100\ \Rightarrow\ n+1>10\ \Rightarrow\ n\ge10$$

take the positive root; $n+1=10$ gives exactly $0.01$, so we need strictly more and land on $n=10$

Confirm the achieved bound
$$b_{11}=\tfrac1{121}\approx0.0083<0.01$$

at $n=10$ the first dropped term is already under the target, so ten terms suffice

Answer $$\boxed{n=10}$$
Check

At $n=9$, $b_{10}=1/100=0.01$, not strictly below, so $10$ is the smallest that works.

Alternating error is free to estimate: pick $n$ so the first dropped term clears your tolerance, then check the boundary case is strict.

D · interleaved 3 questions
1§03.1 — sequence versus series●●○○○

Two questions that students routinely merge. For $a_n=\frac{n}{n+1}$, consider both the sequence $\{a_n\}$ and the series $\sum a_n$.

Given

  • $a_n=\frac{n}{n+1}$

Find

  1. (a) Which is true?

Hint 1/4

Keep the two objects apart: a sequence has a limit; a series has a sum. They are different questions.

Hint 2/4

The sequence limit is $\lim a_n$; the series needs the n-th term test first: if $a_n\not\to0$, $\sum a_n$ diverges.

Hint 3/4

Here $a_n=\frac{n}{n+1}\to1$, so restate: the sequence tends to $1$, and the terms do not vanish.

Hint 4/4

So the sequence converges (to $1$) while the series diverges (terms miss zero).

Show solution

We answer the sequence and the series separately, because they ask different questions; the n-th term test links them but does not merge them.

Find the sequence limit
$$\lim_{n\to\infty}\frac{n}{n+1}=1$$

dividing top and bottom by $n$ leaves $1/(1+1/n)\to1$, so the sequence settles

Test the series on that limit
$$a_n\to1\neq0\ \Rightarrow\ \sum a_n\ \text{diverges}$$

a series whose terms do not vanish cannot converge, so the n-th term test rejects it

Answer $$\boxed{\text{seq.}\to1,\ \text{series diverges}}$$
Check

A series can only converge if its terms vanish; these settle at $1$.

Sequence limit and series sum are different objects; a sequence can settle at a nonzero value while its series has no hope of converging.

2§03.2 — splitting a mixed series●●○○○

A series that is the sum of two familiar pieces. Recall from earlier that a series splits across a sum only when the pieces themselves behave.

Given

  • $\sum_{n=1}^\infty \left(\left(\tfrac12\right)^n+\tfrac1n\right)$

Find

  1. (a) Converge or diverge?

Hint 1/4

Look at the two pieces separately: one is geometric, one is the harmonic series.

Hint 2/4

A sum of series converges only if both pieces converge; if one diverges, so does the sum.

Hint 3/4

The geometric $\sum(1/2)^n$ converges, but $\sum 1/n$ diverges; restate: convergent plus divergent.

Hint 4/4

A convergent plus a divergent series diverges, so the whole thing diverges.

Show solution

We split the sum and judge each piece on its own, valid here because a convergent series added to a divergent one is forced to diverge.

Classify the geometric piece
$$\sum(1/2)^n\ \text{converges}$$

ratio $1/2<1$, so this half totals a finite amount

Classify the harmonic piece
$$\sum 1/n\ \text{diverges}$$

the harmonic series, $p=1$, has no finite total

Combine the two fates
$$\text{finite}+\infty=\infty\ \Rightarrow\ \text{diverges}$$

adding a finite total to an unbounded one stays unbounded, so the combined series diverges

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

If the sum converged, subtracting the convergent geometric part would make $\sum 1/n$ converge, which it does not.

Convergent plus divergent is always divergent; only when both pieces converge may you split and add the sums.

3§03.5 — pick the test yourself●●●○○

No test is named and the type is hidden. Read the term and choose.

Given

  • $\sum_{n=1}^\infty (-1)^{n-1}\,\frac{n^2}{n^3+1}$

Find

  1. (a) Converge or diverge?

Hint 1/4

First name what kind of series this is; the alternating sign is the loudest clue.

Hint 2/4

For an alternating series, test the sizes $b_n=\frac{n^2}{n^3+1}$: do they decrease to zero?

Hint 3/4

$b_n\approx\frac{n^2}{n^3}=\frac1n\to0$ and eventually decreasing; restate: sizes decrease to $0$.

Hint 4/4

So the Alternating Series Test gives convergence.

Show solution

The $(-1)^{n-1}$ flags an alternating series, so we check the size conditions rather than comparing the signed term to a p-series, which the comparison tests forbid.

Check the sizes vanish
$$b_n=\frac{n^2}{n^3+1}\approx\frac1n\to0$$

the leading powers give a $1/n$ rate, so the sizes reach zero

Check the sizes decrease
$$b_n\ \text{decreasing for}\ n\ge1$$

the denominator outgrows the numerator, so each size falls; a few early terms never affect convergence

Fire the test
$$b_n\downarrow0\ \Rightarrow\ \text{converges}$$

both Leibniz conditions hold, so the alternating series converges

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

The signless sizes behave like $1/n$ and diverge, so again the signs carry the convergence.

When no test is named, let the $(-1)^{n-1}$ pick the tool: an alternating series is tested on its sizes, never compared as if it were positive.

Mistake ledger (12 entries)
⚠ Reporting the integral as the sum

the integral is the number you just computed, so it is tempting to hand it in as the total

wrong$$\sum \tfrac1{n^2}=\int_1^\infty x^{-2}\,dx=1$$
right$$\sum \tfrac1{n^2}=\tfrac{\pi^2}{6}\approx1.6449$$
⚠ Skipping the decreasing check

positivity is obvious, so the decreasing hypothesis gets assumed rather than shown

wrong$$a_n>0\ \Rightarrow\ \text{Integral Test applies}$$
right$$f>0\ \text{and}\ f'<0\ \text{on}\ [N,\infty)$$
⚠ Making the cutoff non-strict

$p=1$ feels like it should be included with the converging cases next to it

wrong$$\sum \tfrac1{n^p}\ \text{converges for}\ p\ge1$$
right$$\sum \tfrac1{n^p}\ \text{converges for}\ p>1$$
⚠ Leaving the term as a product of powers

$\frac1{n\sqrt n}$ does not look like a p-series until the powers are combined

wrong$$\tfrac1{n\sqrt n}\ \text{is not a p-series}$$
right$$\tfrac1{n\sqrt n}=n^{-3/2},\ p=\tfrac32$$
⚠ Comparing the wrong way

any true inequality feels like progress, even one that points at the useless yardstick

wrong$$a_n\le \tfrac1n,\ \sum\tfrac1n\ \text{div.}\ \Rightarrow\ \sum a_n\ \text{div.}$$
right$$a_n\le b_n,\ \sum b_n\ \text{conv.}\ \Rightarrow\ \sum a_n\ \text{conv.}$$
⚠ Ignoring that terms must be positive

the test is stated for sizes, and signs get overlooked

wrong$$\text{apply comparison to}\ \sum \tfrac{(-1)^n}{n^2}$$
right$$\text{comparison needs}\ a_n,b_n>0$$
⚠ Concluding from a ratio of 0 or infinity

any limit feels like a usable answer, but only $(0,\infty)$ gives the two-way conclusion

wrong$$\tfrac{a_n}{b_n}\to0\ \Rightarrow\ \text{same fate}$$
right$$\text{need}\ 0<\lim\tfrac{a_n}{b_n}<\infty$$
⚠ Picking the wrong dominant power

keeping a lower-order term in the estimate points at the wrong p-series

wrong$$\frac{n+2}{n^3+1}\approx\frac1n$$
right$$\frac{n+2}{n^3+1}\approx\frac1{n^2}$$
⚠ Assuming alternating means convergent

the sign pattern looks like the whole hypothesis

wrong$$\text{alternating}\ \Rightarrow\ \text{converges}$$
right$$\text{need}\ b_n\downarrow\ \text{and}\ b_n\to0$$
⚠ Checking decrease but not vanishing

the decreasing condition is the fussier one, so it grabs the attention

wrong$$b_{n+1}\le b_n\ \Rightarrow\ \text{converges}$$
right$$b_{n+1}\le b_n\ \text{and}\ b_n\to0$$
⚠ Using the alternating bound on a positive series

$b_{n+1}$ is the easiest bound to write, so it gets used everywhere

wrong$$\sum\tfrac1{n^2}:\ \lvert R_n\rvert\le \tfrac1{(n+1)^2}$$
right$$\sum\tfrac1{n^2}:\ R_n\le\int_n^\infty x^{-2}dx=\tfrac1n$$
⚠ Bounding by the last term kept

off-by-one: the term you just added feels like the relevant one

wrong$$\lvert R_n\rvert\le b_n$$
right$$\lvert R_n\rvert\le b_{n+1}$$
Formula card
Theorem 11.3.1: The Integral Test
$$\boxed{\ \sum_{n=1}^{\infty} a_n \ \text{converges}\iff\ \int_{1}^{\infty} f(x)\,dx \ \text{converges}\ }$$

$f$ is continuous, positive, and decreasing on $[1,\infty)$; the terms come from that function: $a_n=f(n)$; these need hold only for $n\ge N$, since the first few terms never change convergence

Theorem 11.3.2: The p-series test
$$\boxed{\ \sum_{n=1}^{\infty}\frac1{n^{p}}\ \text{converges}\iff\ p>1\ }$$

the term is a single power: $a_n=\dfrac1{n^p}$; $p$ is a real constant

Theorem 11.4.1: The Comparison Test
$$\boxed{\ a_n\le b_n:\quad \textstyle\sum b_n\ \text{conv.}\Rightarrow \sum a_n\ \text{conv.};\quad \sum a_n\ \text{div.}\Rightarrow \sum b_n\ \text{div.}\ }$$

both series have positive terms: $a_n,b_n>0$; the comparison holds eventually: $a_n\le b_n$ for all $n\ge N$

Theorem 11.4.2: The Limit Comparison Test
$$\boxed{\ \lim_{n\to\infty}\frac{a_n}{b_n}=c\in(0,\infty):\quad \textstyle\sum a_n\ \text{and}\ \sum b_n\ \text{share a fate}\ }$$

both series have positive terms: $a_n,b_n>0$; the ratio has a positive finite limit: $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=c,\ 0<c<\infty$

Theorem 11.5.1: The Alternating Series Test
$$\boxed{\ b_n\downarrow 0\ \Rightarrow\ \sum_{n=1}^\infty(-1)^{n-1}b_n\ \text{converges}\ }$$

the series alternates: $\sum(-1)^{n-1}b_n$ with $b_n>0$; the sizes decrease: $b_{n+1}\le b_n$ eventually; the sizes vanish: $\displaystyle\lim_{n\to\infty}b_n=0$

Two remainder bounds
$$\boxed{\ \int_{n+1}^{\infty}\! f\,dx\ \le\ R_n\ \le\ \int_{n}^{\infty}\! f\,dx\ }\qquad\boxed{\ \lvert R_n\rvert\le b_{n+1}\ }$$

integral remainder: the Integral Test hypotheses hold for $f$; alternating remainder: the Alternating Series Test hypotheses hold

Check yourself

Close the page and write, from memory: the three hypotheses of the Integral Test, the p-series cutoff (with the right inequality), the two directions of the Direct Comparison Test, the ratio condition of the Limit Comparison Test, the two size conditions of the Alternating Series Test, and the two remainder bounds. Then check against the formula card and mark only the ones you could not produce.

  • State the three Integral Test hypotheses and run it on $\sum 1/(n\ln n)$?

    c-integral-test

  • Say the p-series verdict for $p=1$, $p=\tfrac32$, and $p=0.9$ without hesitation?

    c-p-series

  • Pick a comparison that points the right way for a convergence proof and for a divergence proof?

    c-comparison

  • Name the yardstick and take the ratio limit for $\frac{2n^2+3n}{\sqrt{5+n^5}}$?

    c-limit-comparison

  • Check both alternating conditions and know which failure sends you to the n-th term test?

    c-alternating

  • Use $\lvert R_n\rvert\le b_{n+1}$ to count the terms for a target accuracy?

    c-estimates

Glossary (11 terms)
infinite seriesseri

The sum $a_1+a_2+a_3+\cdots$ of the terms of a sequence, defined as the limit of its partial sums when that limit exists.

partial sumkısmi toplam

The finite total $s_N=a_1+\cdots+a_N$ of the first $N$ terms; the series converges exactly when these have a finite limit.

p-seriesp serisi

A series $\sum 1/n^p$ for a constant $p$; it converges if and only if $p>1$.

harmonic seriesharmonik seri

The p-series $\sum 1/n$ with $p=1$; the standard example of a divergent series whose terms tend to zero.

Integral Testintegral testi

A test that ties a positive decreasing series to the improper integral of its matching function, so the two share convergence.

Comparison Testkarşılaştırma testi

A test that decides a positive series by a term-by-term inequality against a series of known behaviour.

Limit Comparison Testlimit karşılaştırma testi

A test that links two positive series when the limit of their ratio is a positive finite number, so they share a fate.

alternating seriesalmaşık seri

A series whose terms switch sign, written $\sum(-1)^{n-1}b_n$ with $b_n>0$.

remainderkalan

The tail $R_n=S-s_n$ left after $n$ terms; a bound on it turns a partial sum into a guaranteed estimate.

convergentyakınsak

Describing a series whose partial sums approach a finite limit.

divergentıraksak

Describing a series whose partial sums have no finite limit.

What comes next
§04 · Absolute convergence, the Ratio and Root Tests, and power series

This section decides positive series and cleanly alternating ones. Next comes the tool for the messy middle: absolute convergence, and the Ratio and Root Tests that read a series' fate from a single limit, followed by power series, where these tests find a radius of convergence.

Sources
  • James Stewart, Calculus, ninth edition — sections 11.3, 11.4 and 11.5 The three sections on this week's syllabus line; the theorem numbering and hypotheses follow this book.
  • Standard convergence tests for infinite series The Integral, Comparison, Limit Comparison, and Alternating Series Tests are standard results, stated here in the form used across calculus courses.

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