6 concepts17 worked examples27 exercises4 exam-level5 figures
What are you here for?
04Absolute convergence, the ratio and root tests, a strategy for testing series, and power series
A spreadsheet estimates $\ln 1.5$ by adding $0.5-\frac{0.5^2}{2}+\frac{0.5^3}{3}-\cdots$, and by the sixth row it has settled on $0.405$ and stopped moving. Feed the same column $\ln 3$ as $2-\frac{2^2}{2}+\frac{2^3}{3}-\cdots$ and the running total swings wider every row and never lands. Identical recipe, one number in place of another.
By the end of this section you can decide whether an infinite sum settles even when its signs are irregular, pick the one test out of six that cracks a given series in a single line, and say exactly which inputs a accepts and which it throws out.
In 60 seconds
Two tests read the ratio or the $n$th root of the terms and compare the series to a geometric one; one flowchart decides which test to reach for; and the payoff is that a power series is a geometric series with a radius, trustworthy inside it and untouched outside.
Reading $L=1$ from the ratio or root test as a verdict. It is the test resigning: $\sum \frac1n$ and $\sum \frac1{n^2}$ both give $L=1$, yet one diverges and one converges.
Dropping the absolute value in the ratio test. The test compares $\left|a_{n+1}/a_n\right|$; a stray minus sign has made many a limit of $-2$ look like it passes $L<1$.
Naming the interval of convergence without testing the two endpoints. The ratio test is silent exactly at $|x-a|=R$, so each endpoint is a separate series to settle by hand.
Quizzes and homework are each worth $10\%$ of the grade; the syllabus lists no other weights, so on the evidence given no single topic here is the exam's known favourite. Radius-of-convergence and ratio-test questions are short to state and to grade, which is worth knowing when the clock is part of the score.
How much time do you have?
10 minutes
You leave able to run the ratio test, spot when $L=1$ means try something else, and find a radius of convergence. Enough to survive a short test this series or find $R$ question, not a full interval-with-endpoints problem.
The 60 second card, Formula card, The ratio test: compare the series to a geometric one, Mistake ledger
45 minutes
You add the two things the formulas cannot supply on their own: how to choose a test from the shelf of six, and how to pin down the two endpoints of an interval of convergence.
The 60 second card, Absolute against conditional convergence, The ratio test: compare the series to a geometric one, A strategy: reading a series before testing it, Radius and interval of convergence, Worked exam question
full read
Every block, including the root test, the fading-scaffold ladder, and the interleaved practice that forces the choose-a-test decision on you without a heading to give the answer away.
read top to bottom
By the end of this section
Distinguish absolute from conditional convergence, and use the theorem that absolute convergence forces ordinary convergence on a series whose signs are irregular.
Apply the ratio test, read its limit correctly, and recognise when the outcome is no verdict.
Apply the root test to a term that is an expression raised to the power n, and relate its verdict to the ratio test.
Choose the cheapest correct test for a series presented with no hint, starting from the term-going-to-zero check.
Read off the centre and coefficients of a power series and state the convergence trichotomy it must obey.
Find the radius of convergence with the ratio test and complete the interval by testing both endpoints.
Syllabus coverage
11.6
Absolute convergence and the ratio and root tests
Split across three blocks: the absolute-convergence theorem, then the ratio test and the root test, each with its own proof idea and its own failure at L equal to one.
covered
11.7
Strategy for testing series
Written as a decision procedure with the flowchart drawn out, because 11.7 adds no new test; its whole content is the order in which you try the tests you already have.
covered
11.8
Power series
The definition and the convergence trichotomy sit in one block; computing the radius and completing the interval by testing endpoints sit in the next.
covered
Recall first
Geometric series
$\sum_{n=0}^{\infty} a r^{n}$ converges exactly when $|r|<1$, and then its sum is $\dfrac{a}{1-r}$. Every test in this section works by comparing the given series to a geometric one.
The ratio and root tests both prove convergence by trapping the terms under a geometric series with ratio below one.
p-series
$\sum_{n=1}^{\infty}\dfrac{1}{n^{p}}$ converges exactly when $p>1$. In particular $\sum \frac1n$ diverges and $\sum \frac1{n^2}$ converges.
Endpoints of an interval of convergence almost always turn into a p-series or an alternating p-series.
Test for divergence (the n-th term test)
If $\lim_{n\to\infty} a_n \neq 0$, or the limit does not exist, then $\sum a_n$ diverges. It can never prove convergence.
It is the first line of the testing strategy and the cheapest check you can run.
Alternating series test
$\sum (-1)^{n-1} b_n$ with $b_n>0$ converges provided $b_n$ decreases and $b_n\to 0$. The alternating harmonic series $\sum \frac{(-1)^{n-1}}{n}$ is the standard example.
It is what settles the convergent endpoint of many intervals, and it produces the standard series.
Two limits that keep appearing
$\lim_{n\to\infty}\left(1+\frac1n\right)^{n}=e$ and $\lim_{n\to\infty}\sqrt[n]{n}=1$ and $\lim_{n\to\infty}\sqrt[n]{c}=1$ for any constant $c>0$.
They are exactly the limits that come out of the ratio and root tests on the worked examples here.
Try it yourself first (3 questions)
1§04 — an alternating series that does not converge●●○○○
A quick trap before the section starts. A series whose terms flip sign every step looks like a job for the alternating series test, but that test has a hypothesis, and one of the two series below fails it.
Given
the series $\sum_{n=1}^{\infty}(-1)^n$, whose terms are $-1,+1,-1,+1,\dots$
Find
(a) True or false: this series converges because it alternates.
Hint 1/4
The alternating series test is not a licence for every series with a $(-1)^n$ in it; check its hypotheses before you invoke it.
Hint 2/4
The test requires $b_n\to 0$. Here $b_n=1$ for every $n$, so ask what the term test says first.
Hint 3/4
The terms are $(-1)^n$, which do not approach $0$: they sit at $-1$ and $+1$ forever. The partial sums bounce between $-1$ and $0$.
Hint 4/4
Since the terms do not go to zero, the term test forces divergence; alternating or not, it cannot converge.
Show solutionCheck the cheapest test first
$$\lim_{n\to\infty}(-1)^n$$
the term test comes before any alternating-series argument, because it is cheaper and can kill the series outright
$$\text{does not exist}$$
the values sit at $-1$ and $+1$ and never approach a single number
Read the verdict
$$\sum(-1)^n\ \text{diverges}$$
a series whose terms do not tend to $0$ cannot converge, whatever its sign pattern
Answer $$\sum(-1)^n\ \text{diverges}$$
Check
The partial sums are $-1,0,-1,0,\dots$, which never settle, confirming divergence directly.
The word alternating on its own proves nothing; the alternating series test needs the terms to shrink to zero.
2§04 — a geometric series you should recognise on sight●○○○○
A geometric series with a ratio you can read off. Recognising these instantly is half of the testing strategy, so warm it up now.
the series is geometric with $|r|<1$, so the closed form applies
$$=\frac{1}{\tfrac13}=3$$
arithmetic
Answer $$3$$
Check
Partial sums $1,\tfrac53,\tfrac{19}{9},\dots$ climb toward $3$ and never pass it, as a convergent geometric series should.
Geometric and p-series are the two families you settle by recognition, not by a test.
3§04 — a p-series check●○○○○
One more recognition check, this time a p-series. The exponent decides everything.
Given
the series $\sum_{n=1}^{\infty}\dfrac{1}{n^{3/2}}$
Find
(a) True or false: this series converges.
Hint 1/4
Do not test this; classify it. What family does $1/n^{p}$ belong to, and what is $p$ here?
Hint 2/4
A p-series $\sum 1/n^{p}$ converges exactly when $p>1$.
Hint 3/4
Here $p=\tfrac32$, and the data you need is just that one number.
Hint 4/4
Since $\tfrac32>1$, the series converges.
Show solutionRead the exponent
$$p=\tfrac32>1$$
the only thing a p-series verdict depends on is whether $p>1$
$$\Rightarrow\ \text{converges}$$
the p-series rule
Answer $$\text{converges}$$
Check
It sits between $\sum 1/n$ (diverges) and $\sum 1/n^2$ (converges); with $p=1.5$ it lands on the convergent side.
Keep the p-series boundary at $p=1$ memorised; endpoints of power-series intervals land on it constantly.
Notation
symbol
reads as
means
watch out
$\sum a_n$
the sum of a-sub-n
shorthand for $\sum_{n=1}^{\infty} a_n$ (or from whatever index is written), the infinite sum of the terms $a_n$.
The starting index rarely matters for convergence; $\sum_{n=0}$ and $\sum_{n=1}$ of the same rule converge together.
$|a_n|$
the absolute value of a-sub-n
the size of the term with its sign stripped; the series $\sum |a_n|$ is the positive-term series attached to $\sum a_n$.
$\sum a_n$ and $\sum |a_n|$ can have different verdicts; that gap is the whole point of conditional convergence.
$n!$
n factorial
the product $1\cdot 2\cdot 3\cdots n$, with $0!=1$. It eventually outgrows every $c^{n}$.
The cancellation $\dfrac{(n+1)!}{n!}=n+1$ is the move that makes the ratio test love factorials; do not expand the factorials.
$\sum c_n (x-a)^n$
the sum of c-sub-n times x-minus-a to the n
a power series centred at $a$ with coefficients $c_n$; feeding a number in for $x$ turns it into an ordinary series of numbers.
The $c_n$ are constants and $x$ is the variable; the radius $R$ is decided by the $c_n$ alone, not by $x$.
Conventions used here
L equal to one is silence, not a verdict
When the ratio or root test returns $L=1$, the test has failed to decide and you must reach for another one. It never means the series diverges and never means it converges: $\sum \frac1{n^2}$ (converges) and $\sum \frac1n$ (diverges) both return $L=1$.
The ratio test uses absolute values
The ratio test limit is $\lim\left|a_{n+1}/a_n\right|$, with the bars. Whenever a term carries a sign, such as $(-2)^n$, strip the sign before taking the limit, or a negative limit will masquerade as a value below one.
Convergence ignores finitely many terms
Dropping, adding, or changing the first few terms of a series never changes whether it converges; it only changes the sum. This is why a test only has to hold eventually, and why a series that starts at $n=0$, $n=1$, or $n=2$ has the same convergence verdict.
A radius is a number, an interval is a set
The radius of convergence $R$ is a single number the ratio test hands you. The interval of convergence is the full set of $x$ that work, which equals the open interval $(a-R,a+R)$ together with whichever of the two endpoints pass their own separate test. The two endpoints can disagree.
4.1Absolute against conditional convergence
Bounds the size of a messy-signed series so a positive-term test can settle it, and names the two ways a series can converge.
Last week every test but one carried the same fine print — the terms must be positive — and the alternating series test wanted a strict $+,-,+$ pattern. Most series are neither.
Solvable with what we have
positive-term series such as $\sum \frac1{n^2+1}$, by comparison or the integral test
strictly alternating series such as $\sum \frac{(-1)^{n-1}}{n}$, by the alternating series test
Not solvable yet
$\sum \frac{\sin n}{n^2}$, whose signs follow no fixed pattern
$\sum \frac{\cos n}{n^2}$, positive sometimes and negative other times, with no regularity
Try the comparison test on $\sum \frac{\sin n}{n^2}$: it needs $0\le a_n\le b_n$, but $\frac{\sin n}{n^2}$ is negative for many $n$, so there is nothing to bound below. Try the alternating series test: it needs a clean $(-1)^{n-1}b_n$ shape with $b_n$ decreasing, but $\sin n$ does not alternate on a fixed schedule, so there is no $b_n$ to decrease.
Why it fails
Both tools were built for a sign pattern this series simply does not have. We need a way to handle any signs at all.
If the series of sizes converges, so does the original series. We call the original series when $\sum|a_n|$ converges; we call it conditionally convergent when it converges but $\sum|a_n|$ diverges, so its convergence leans entirely on cancellation between positive and negative terms.
The running total of $\textcolor{#d1690a}{\sum|a_n|}$ only ever climbs, and it stops at a ceiling. The running total of $\textcolor{#1f6feb}{\sum a_n}$ moves both ways but can never escape the band the ceiling sets, so it too is cornered into a limit.
Looks like this, but is not
$\sum \frac{(-1)^{n-1}}{n}$ converges, so surely the theorem applies and $\sum \frac1n$ converges too.
The theorem runs one way only. Here $\sum \frac1n$ is the harmonic series and diverges, so the hypothesis fails and the theorem says nothing. This series is only conditionally convergent, held up by cancellation, not by fast enough shrinking.
A series with irregular signs: sin n over n squared
The series that defeated both of last week's applicable tests, settled in three lines.
Squaring the denominator is the only change, and it turns the divergent harmonic tail into a convergent one; that is precisely what moves a series from conditional to absolute.
Same $(-1)^{n-1}$, different exponent, different category. The category is read off the absolute series alone.
Checkpoint
§04.1 — using the theorem backwards●●○○○
A thirty second check on the direction of the theorem. The implication points one way; the reverse is a trap that catches many.
Given
the claim: if $\sum |a_n|$ diverges, then $\sum a_n$ diverges
Find
(a) True or false?
Hint 1/4
Ask whether one convergent series you already know breaks this claim.
Hint 2/4
The theorem only says converge implies converge; it says nothing when $\sum|a_n|$ diverges.
Hint 3/4
Take $a_n=\frac{(-1)^{n-1}}{n}$: then $\sum|a_n|=\sum\frac1n$ diverges, yet $\sum a_n$ converges by the alternating series test.
Hint 4/4
So a divergent absolute series does not force divergence; the claim is false.
right$$\left|\frac{\sin n}{n^2}\right|\le \frac1{n^2}\ \text{compare the sizes}$$
4.2The ratio test: compare the series to a geometric one
Reads the limit of $|a_{n+1}/a_n|$; below one the series converges, above one it diverges, at one it says nothing.
The absolute-convergence theorem turns any series into a positive-term one, so now we want a positive-term test that fires exactly where comparison and the integral test choke: on factorials and on constants raised to the power $n$.
TheoremTheorem 11.6.5: the ratio test
Conditions
the limit $L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$ exists or is $+\infty$
If eventually each term is less than a fixed fraction of the one before, the terms fall at least as fast as a geometric series and the sum converges. If each term outgrows the one before, the terms cannot even shrink to zero, so the series diverges.
Once $\textcolor{#1f6feb}{|a_n|}$ shrinks by at least a fixed factor $r<1$ each step, every bar sits under the $\textcolor{#d1690a}{\text{geometric envelope } r^{\,n}}$. A convergent geometric series is holding a roof over the whole tail.
Looks like this, but is not
The ratio test on $\sum \frac1{n^2}$ gives $\lim \frac{n^2}{(n+1)^2}=1$, which feels like it ought to settle the matter.
$L=1$ settles nothing. This series converges, but as a p-series, not through the ratio test. On $\sum \frac1n$ the ratio test also returns $L=1$, and that one diverges, so $L=1$ genuinely carries no information.
A factorial in the denominator: 2 to the n over n factorial
The kind of series comparison cannot easily reach, cracked by the ratio test in two lines.
When the whole term is an expression to the power $n$, the $n$th root turns it straight into a limit to compare with one.
The ratio test loves factorials because each factorial nearly cancels the next. When the whole term is instead something raised to the power $n$, a cleaner cancellation is available: take the $n$th root.
TheoremTheorem 11.6.6: the root test
Conditions
the limit $L=\lim_{n\to\infty}\sqrt[n]{|a_n|}$ exists or is $+\infty$
Take the $n$th root of the size of the $n$th term. If that root settles below one, the terms are eventually smaller than $r^n$ for some $r<1$, a convergent geometric series; if it settles above one, the terms cannot go to zero.
Looks like this, but is not
$\sum\left(\frac{n}{2n+1}\right)^{n}$ has an $n$ in it, so the ratio test looks mandatory.
The ratio test here is a tangle of $n$th powers. The root test is one line: $\sqrt[n]{|a_n|}=\frac{n}{2n+1}\to\frac12<1$. When the whole term is raised to the $n$, take the root, not the ratio.
A ratio of linear terms, all raised to the n
The natural home of the root test: the exponent is $n$ and nothing else is going on.
The worked example $\sum n!/n^n$ gave $1/e$ by both routes, a concrete instance of the agreement.
Choose the test by which limit is easier, not by which answer you want; they match.
⚠ The n-th root of n sent to zero
an $n$th root feels like it should crush anything, including a lone factor of $n$
wrong$$\sqrt[n]{n}\to 0$$
right$$\sqrt[n]{n}\to 1$$
⚠ Rooting only part of the term
the power $n$ is visible on one piece, so the root gets applied there only
wrong$$\sqrt[n]{\frac{2^n}{n^n}}=\frac{2^n}{n}$$
right$$\sqrt[n]{\frac{2^n}{n^n}}=\frac{2}{n}$$
4.4A strategy: reading a series before testing it
Adds no new test; it is the order in which you try the six you already have, cheapest and most decisive first.
Six tests now sit on the shelf. The exam will not tell you which one to reach for, and reaching for the wrong one first wastes the minutes you are scored on. The whole content of 11.7 is the order.
MethodStrategy 11.7: choose a test by the shape of the term
Conditions
you have a series $\sum a_n$ and no instruction on which test to use
Before any real work, glance at the term itself: if it does not shrink to zero, the series is dead on arrival and you are done. Only if the terms do go to zero do you spend effort choosing among the convergence tests.
Read the term, not the whole series, and walk the chart top to bottom. The first box is free and kills more exam series than any other single check; the shape of the term picks the branch after that.
Looks like this, but is not
$\sum \frac{1}{n^2+1}$ has an $n$ growing inside it, so the ratio test looks like the move.
The ratio test gives $L=1$ here and wastes your time. The term looks like $\frac1{n^2}$, so a limit comparison with the convergent p-series $\sum \frac1{n^2}$ settles it in one line. Match the term to a family first.
Four series, four different first moves
Not a computation but a sorting exercise: for each series, name the test the chart points to and the one-line reason, without grinding through it.
Given
(A) $\sum \dfrac{n^2-1}{n^3+1}$
(B) $\sum \dfrac{n!}{e^{n}}$
(C) $\sum \dfrac{(-1)^n}{\ln(n+1)}$
(D) $\sum \dfrac{3^n}{n^3}$
Find
the cheapest correct test for each
SolutionA: a rational function of n
$$\frac{n^2-1}{n^3+1}\sim \frac{1}{n}$$
leading behaviour is $1/n$, so limit-compare with the divergent $\sum 1/n$
A series with a dial: feed a number in for $x$ and it becomes an ordinary series, converging on the centre, everywhere, or an interval.
Every series so far has been a fixed bag of numbers with one yes or no answer. A power series has a dial on it: put a value in for $x$ and you get an ordinary numerical series back, with its own verdict.
TheoremTheorem 11.8.3: the convergence trichotomy
Conditions
$\sum_{n=0}^{\infty} c_n (x-a)^n$ is any power series centred at $a$, with constant coefficients $c_n$
$$\boxed{\;\text{exactly one holds: (i) converges only at } x=a;\ (\text{ii) for all } x;\ (\text{iii) } |x-a|<R\ \text{conv.},\ |x-a|>R\ \text{div.}\;}$$
For a power series there are only three shapes the set of good inputs can take: a single point (the centre), the whole line, or an interval of some half-width $R$ centred at $a$. The number $R$ is called the radius of convergence, with $R=0$ and $R=\infty$ covering the first two cases.
The only three shapes a power series can converge on: the centre alone $(R=0)$, the whole line $(R=\infty)$, or a $\textcolor{#128a5a}{\text{green interval}}$ of radius $R$ about the centre. Nothing else is possible.
Looks like this, but is not
$\sum n!\,x^n$ has terms that shrink once $x$ is small enough, so it should converge near $0$.
No nonzero $x$ is small enough: the ratio $(n+1)|x|\to\infty$ for every $x\neq 0$. This is the extreme case $R=0$, where the interval of convergence collapses to the single point $x=0$.
A power series that converges nowhere but the centre
The extreme $R=0$ case, where a factorial coefficient strangles every nonzero input.
both $c_n$ and $(x-a)^n$ carry the index $n$, so they blur together
wrong$$c_n=(x-a)^n$$
right$$c_n\ \text{is the constant multiplying}\ (x-a)^n$$
4.6Radius and interval of convergence
The ratio test hands you the radius $R$ and goes silent at the two endpoints, which you then settle one at a time.
The trichotomy promises a radius $R$ but does not hand it over. The ratio test does, and it also refuses to speak at the two endpoints, which is exactly where the interval's brackets are decided.
MethodMethod: radius, then endpoints
Conditions
$\sum c_n(x-a)^n$ with the ratio (or root) test available on $|c_n|$
$$\boxed{\;\text{solve}\ \lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right|\,|x-a|<1\ \Longleftrightarrow\ |x-a|<R,\ \text{then test } x=a\pm R\ \text{separately}\;}$$
Run the ratio test with $x$ carried along; the limit comes out as some constant times $|x-a|$, and setting that below one reads off $R$. The test says nothing at $|x-a|=R$, so plug each endpoint back in and test the resulting number series on its own.
Inside the radius the series $\textcolor{#128a5a}{\text{converges}}$, outside it $\textcolor{#b3271e}{\text{diverges}}$, and the ratio test is silent at the two open circles. Each endpoint is its own number series, tested separately, and the two can disagree.
Looks like this, but is not
$\sum \frac{x^n}{n}$ converges for $|x|<1$ by the ratio test, so its interval of convergence is $(-1,1)$.
The ratio test is silent at $x=\pm 1$, and the two endpoints disagree: $x=1$ gives the divergent harmonic series, $x=-1$ the convergent alternating harmonic series. The honest interval is $[-1,1)$, not $(-1,1)$.
Radius one, and the two endpoints split
The cleanest interval problem, and the one that shows why both endpoints must be tested.
the alternating harmonic series by the AST, so $x=-1$ is included
Assemble the interval
$$[-1,1)$$
closed at the endpoint that converged, open at the one that diverged
Answer $$R=1,\qquad \text{interval } [-1,1)$$
Check
$R=1$ matches the plain geometric series $\sum x^n$; dividing by $n$ left the radius alone and only changed the endpoints, which is precisely where the extra work landed.
One ratio limit, two endpoint series.
The two endpoints are two different series and can disagree; test both, always.
A base of three sets the radius to one third
A coefficient carrying a constant to the n; watch how it fixes both the radius and the sign at each endpoint.
The two endpoints became an alternating p-series (converges) and a plain p-series (diverges), both with $p=\tfrac12$; the $p=\tfrac12$ boundary is exactly why they split.
A base $b^n$ in the coefficient sets $R=1/b$, and the sign it leaves at each endpoint decides which p-series you face.
Checkpoint
§04.6 — a radius from a geometric coefficient●●○○○
A thirty second computation of a radius; no endpoints asked for, just $R$.
the term has a factorial, or a constant raised to the power n, or both.
Write the ratio
Form $\left|\dfrac{a_{n+1}}{a_n}\right|$: put the $(n+1)$-th term over the $n$-th and keep every factorial as a factorial.
Cancel in pieces
Handle each factor on its own: powers give a constant ($c^{n+1}/c^{n}=c$), factorials collapse ($(n+1)!/n!=n+1$), and polynomials group into ratios like $(n+1)/n$.
Take the limit
Let $n\to\infty$ to get $L$. Watch for the standard limits $(1+1/n)^n\to e$ and $(n+1)/n\to 1$.
Read the verdict
$L<1$ converges absolutely, $L>1$ (or $\infty$) diverges, $L=1$ is no verdict, so switch tests.
Where it goes wrong
Dropping the absolute value when a $(-1)^n$ or $(-c)^n$ is present.
Expanding $(n+1)!$ instead of cancelling it against $n!$.
Treating $L=1$ as a divergence verdict.
Choosing a test in order
you are handed a series with no hint which test to run.
Term test first
If $a_n\not\to 0$, stop: the series diverges. This is free and catches many series.
Recognise a family
Is it geometric ($\sum ar^n$) or a p-series ($\sum 1/n^p$)? If so you already know the answer.
Look like a p-series?
A rational function of $n$ behaves like its leading powers; use direct or limit comparison.
A sign that alternates?
A clean $(-1)^{n}$ points to the alternating series test, after you check whether it converges absolutely.
Factorial or constant to the n?
Reach for the ratio test (factorials) or the root test (whole term raised to the $n$).
Otherwise
Fall back on the integral test when the term is a positive decreasing function you can integrate.
Where it goes wrong
Running the ratio test on a rational function of $n$ and getting the useless $L=1$.
Skipping the term test and grinding a full computation on an obviously divergent series.
Radius and interval in four moves
a power series $\sum c_n(x-a)^n$ is given and you need the set of x it converges on.
Ratio test carrying x
Compute $\lim\left|\dfrac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right|$; it comes out as (constant) times $|x-a|$.
Read off R
Set that limit below one and solve for $|x-a|$; the boundary value is the radius $R$.
Test both endpoints
Substitute $x=a-R$ and $x=a+R$ one at a time; each gives an ordinary number series to settle.
Bracket the interval
Close the bracket at each endpoint that converged, leave it open at each that diverged.
Where it goes wrong
Stopping at the open interval and never testing the endpoints.
Assuming the two endpoints must agree.
Absolutely convergent: alternating over n squared
An alternating series whose absolute version already converges.
Line 1: the powers give $3^{n+1}/3^{n}=3$, and the factorials give $n!/(n+1)!=1/(n+1)$.
$\lim_{n\to\infty}\dfrac{3}{n+1}=0$
reasoning
Line 2: the denominator grows without bound, so the ratio collapses to $0$.
$L=0<1$, so the series converges.
reasoning
Line 3: $L=0$ is below $1$, so the ratio test returns convergence.
3 · find the buried error
Harder, and now with two errors buried in a plausible solution. Read $\sum \dfrac{(-2)^n}{n^2+1}$ tested by the ratio test, and find the two steps that are wrong.
Step 2. Leave it as $\dfrac{-2\,(n^2+1)}{(n+1)^2+1}$.
Step 3. Since $\dfrac{n^2+1}{(n+1)^2+1}\to 0$, the limit is $-2\cdot 0=0$.
Step 4. $L=0<1$, so the series converges.
the two buried errors (2)
⚠ step 1
the absolute value was dropped. The ratio test uses $\left|a_{n+1}/a_n\right|$, so the factor $(-2)$ must enter as $+2$, not $-2$.
the sign is carried along out of algebra habit, where you would keep it; the bars are supposed to remove it
right
Take sizes: $\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{2\,(n^2+1)}{(n+1)^2+1}$.
⚠ step 3
the rational factor tends to $1$, not $0$. Numerator and denominator both have degree two with leading coefficient $1$, so the ratio tends to the ratio of leading coefficients.
a bigger-looking denominator usually kills a fraction, but here the degrees match, so the limit is $1$
right
$\dfrac{n^2+1}{(n+1)^2+1}\to 1$, so $L=2\cdot 1=2>1$ and the series diverges.
4 · the bare problem
§04.2 — a bare ratio-test problem●●●○○
No scaffold this time. Decide whether the series converges, using the ratio test.
Given
$\sum_{n=1}^{\infty} \dfrac{n!}{100^{n}}$
Find
(a) Converge or diverge? Justify with the ratio test.
Hint 1/4
A factorial against a constant to the $n$: set up the ratio and see which side wins as $n$ grows.
Hint 2/4
The ratio is $\dfrac{(n+1)!}{100^{n+1}}\cdot\dfrac{100^{n}}{n!}=\dfrac{n+1}{100}$.
Hint 3/4
For $a_n=n!/100^{n}$: the factorials give $n+1$, the powers give $1/100$.
Hint 4/4
As $n\to\infty$, $\dfrac{n+1}{100}\to\infty>1$, so the series diverges.
the powers give $(-1)^n$, leaving the alternating harmonic series
$$\Rightarrow\ \text{converges}$$
alternating harmonic by the AST, so $x=-1$ is in
Assemble the interval
$$[-1,\,5)$$
closed at $-1$ where it converged, open at $5$ where it diverged
Answer $$R=3,\qquad \text{interval } [-1,5)$$
Check
Centre $2$ with radius $3$ gives the raw interval $(-1,5)$; the endpoints turn into the harmonic series (out) and the alternating harmonic series (in), so the honest brackets are $[-1,5)$.
One ratio limit, two endpoint series.
A coefficient $1/(n\,b^n)$ centred at $a$ gives $R=b$ and typically leaves a harmonic / alternating-harmonic split at the two ends.
Practice
A · concept 4 questions
1§04.1 — is every convergent series absolutely convergent●●○○○
A one-line conceptual claim about the two kinds of convergence. One familiar series settles it.
Given
the claim: every convergent series converges absolutely
Find
(a) True or false?
Hint 1/4
Look for a convergent series whose absolute version you know diverges.
Hint 2/4
Absolute convergence means $\sum|a_n|$ converges, a strictly stronger demand than $\sum a_n$ converging.
Hint 3/4
The alternating harmonic series $\sum\frac{(-1)^{n-1}}{n}$ converges, yet $\sum\frac1n$ diverges.
Hint 4/4
So it converges without converging absolutely; the claim is false.
Show solutionExhibit the witness
$$\sum\frac{(-1)^{n-1}}{n}\ \text{converges}$$
alternating series test
$$\sum\frac1n\ \text{diverges}$$
harmonic series
Conclude
$$\text{convergent, not absolutely}\ \Rightarrow\ \text{claim false}$$
one counterexample is enough
Answer $$\text{false}$$
Check
Conditionally convergent series are by definition the counterexamples to this claim.
Convergent is weaker than absolutely convergent; the gap is exactly the conditional case.
2§04.2 — what L equal to one means●●○○○
A claim about the ratio test at its blind spot. Decide it with a pair of series you already know.
Given
the claim: if $\lim\left|a_{n+1}/a_n\right|=1$, the series diverges
Find
(a) True or false?
Hint 1/4
Ask whether a single value of $L$ can occur for both a convergent and a divergent series.
Hint 2/4
$L=1$ is the ratio test's no-verdict case; it forbids no outcome.
Hint 3/4
Both $\sum\frac1{n^2}$ (converges) and $\sum\frac1n$ (diverges) give $L=1$.
Hint 4/4
Since a convergent series also gives $L=1$, the claim is false.
Show solutionShow both verdicts occur at L = 1
$$\sum\frac1{n^2}:\ L=1,\ \text{converges}$$
ratio $\to 1$, but it is a convergent p-series
$$\sum\frac1{n}:\ L=1,\ \text{diverges}$$
ratio $\to 1$, and it is the divergent harmonic series
Conclude
$$\text{claim false}$$
a convergent series shares the value $L=1$
Answer $$\text{false}$$
Check
This is exactly why $L=1$ is called no verdict rather than a divergence result.
Never read $L=1$ as either answer; change tests.
3§04.2 — the ratio test on a p-series●●●○○
You run the ratio test on a convergent p-series. The point is to see what the test does and does not tell you.
Given
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$
Find
(a) What does the ratio test give, and what may you conclude from it?
Hint 1/4
Compute the ratio limit first, then ask what that particular value licenses.
Hint 2/4
The ratio is $\dfrac{n^2}{(n+1)^2}=\left(\dfrac{n}{n+1}\right)^2$.
Hint 3/4
That tends to $1$, which is the ratio test's no-verdict value.
Hint 4/4
So the ratio test gives $L=1$ and concludes nothing here, even though the series converges as a p-series.
$$\sum c_n(x-a)^n:\ \text{conv. only at } a,\ \text{for all } x,\ \text{or on } |x-a|<R$$
constant coefficients, centre a
Radius and interval
$$\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right||x-a|<1\ \Leftrightarrow\ |x-a|<R,\ \text{then test } x=a\pm R$$
ratio test applied to the coefficients
Check yourself
Close the page. On one sheet, from memory: the ratio test and the root test with what $L<1$, $L>1$ and $L=1$ each mean; the one sentence that says why absolute convergence gives ordinary convergence; the first question the testing strategy makes you ask; and the two-step recipe for an interval of convergence. Then open the formula card and mark what was missing. What you missed is your revision list; the rest of this section is not.
state the difference between absolute and conditional convergence and give one example of each?
c-absolute
run the ratio test on a factorial series and say what L equal to one would mean?
c-ratio
apply the root test to a term raised to the power n?
c-root
name the first test to try on an unfamiliar series and why?
c-strategy
read off the centre and coefficients of a power series and state its three possible convergence shapes?
c-power
find a radius of convergence and complete the interval by testing both endpoints?
c-radius
Glossary (7 terms)
absolutely convergentmutlak yakınsak
A series $\sum a_n$ is absolutely convergent when the series of sizes $\sum |a_n|$ converges; such a series always converges.
conditionally convergentkoşullu yakınsak
A series that converges while $\sum |a_n|$ diverges, so its convergence relies on cancellation between positive and negative terms.
ratio testoran testi
A test using $L=\lim|a_{n+1}/a_n|$: the series converges absolutely if $L<1$, diverges if $L>1$, and is undecided if $L=1$.
root testkök testi
A test using $L=\lim\sqrt[n]{|a_n|}$, with the same $L<1$, $L>1$, $L=1$ outcomes as the ratio test.
power serieskuvvet serisi
A series $\sum c_n(x-a)^n$ with constant coefficients $c_n$ and a variable $x$, centred at the number $a$.
radius of convergenceyakınsaklık yarıçapı
The number $R$ such that the power series converges for $|x-a|<R$ and diverges for $|x-a|>R$; it can be $0$ or infinite.
interval of convergenceyakınsaklık aralığı
The full set of $x$ for which a power series converges: the open interval of radius $R$ plus whichever endpoints pass their own test.
What comes next
§05 · Functions as power series, Taylor and Maclaurin series, and their applications
This section could tell you where a power series converges, but never what it converges to. Next comes the payoff: those same power series are the Taylor and Maclaurin series of familiar functions, so that $e^x$, $\sin x$ and $\ln(1+x)$ each become a power series you can differentiate, integrate, and use to compute values by hand.
Sources
James Stewart, Calculus, Ninth Edition — sections 11.6, 11.7 and 11.8 Section numbers are quoted only as they appear on the week 4 line of the course syllabus: 11.6 absolute convergence with the ratio and root tests, 11.7 the strategy for testing series, 11.8 power series.
Course syllabus: assessment weights Quizzes and homework are each weighted 10%; no other weights are listed, so no further exam claim is made here.
Standard convergence tests The ratio, root, comparison, alternating and integral tests are standard results; the geometric-series comparison behind the ratio and root tests is reproduced on this page rather than cited.