Practice / Calculus

Series convergence tests: geometric, divergence, integral, comparison and alternating

Ten problems on infinite series: a geometric series that starts at n = 1, a telescoping sum, the divergence test and what it cannot say about 1/√n, the integral test on 1/(n ln² n) with bounds on the sum, a direct comparison for (2 + cos n)/n², a limit comparison, the alternating harmonic series and conditional convergence, how many terms an alternating error bound needs, the ratio test on n!/nⁿ, and the step-size conditions behind stochastic gradient descent, with worked solutions and the mistakes of starting a geometric sum at the wrong term, reading terms that go to 0 as convergence and comparing with a bigger divergent series.

Before you start

An infinite series adds up infinitely many numbers, and the first question is whether the total is a finite number at all. Taylor series, the geometric sums behind an exponential moving average, and the step-size conditions in proofs that stochastic gradient descent converges all depend on the answer. A handful of tests settle most series: compare with a geometric series or a $p$ -series, compare with an integral, or use the alternating signs. The work goes wrong in a few places: a geometric sum started at the wrong term, terms that go to $0$ read as proof of convergence, a comparison made in the direction that proves nothing, the alternating series test used on terms that do not shrink, and a ratio limit of $1$ treated as a verdict. These ten problems cover each of them.

Convergence. $\displaystyle\sum_{n=1}^\infty a_n$ converges to $S$ if the partial sums $S_N = a_1 + \dots + a_N$ tend to $S$ Otherwise it diverges.
Geometric series. $\displaystyle\sum_{n=0}^\infty ar^n = \frac{a}{1 - r}$ for $\lvert r\rvert < 1$ and it diverges for $\lvert r\rvert \ge 1$ In general the sum is $\dfrac{\text{first term}}{1 - r}$
Divergence test. If $a_n \not\to 0$ the series diverges. If $a_n \to 0$ this test says nothing.
Integral test and $p$ -series. If $f$ is positive and decreasing with $f(n) = a_n$ then $\sum a_n$ and $\int_1^\infty f(x)\,dx$ converge or diverge together, and $\displaystyle\int_N^\infty f \le \sum_{n=N}^\infty a_n \le a_N + \int_N^\infty f$ So $\displaystyle\sum\frac1{n^p}$ converges exactly when $p > 1$
Comparison. For $0 \le a_n \le b_n$ if $\sum b_n$ converges so does $\sum a_n$ if $\sum a_n$ diverges so does $\sum b_n$ Limit comparison: if $a_n, b_n > 0$ and $a_n/b_n \to L$ with $0 < L < \infty$ both converge or both diverge.
Alternating series. If $b_n$ decreases to $0$ $\sum(-1)^{n+1}b_n$ converges, and stopping after $N$ terms leaves an error of at most $b_{N+1}$ A series is absolutely convergent if $\sum\lvert a_n\rvert$ converges, and conditionally convergent if it converges but $\sum\lvert a_n\rvert$ does not.
Ratio test. If $\lvert a_{n+1}/a_n\rvert \to \ell$ the series converges absolutely for $\ell < 1$ and diverges for $\ell > 1$ At $\ell = 1$ it says nothing.

Builds on: Improper integrals: infinite limits, infinite integrands and comparison, Evaluating limits: factoring, conjugates and limits at infinity

Problems

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Find $\displaystyle\sum_{n=1}^\infty 5\Big(\frac23\Big)^n$
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Find $\displaystyle\sum_{n=1}^\infty \frac{1}{n(n + 1)}$ from its partial sums.
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What does the divergence test say about $\displaystyle\sum_{n=1}^\infty\frac{n}{2n + 1}$ and about $\displaystyle\sum_{n=1}^\infty\frac1{\sqrt n}$
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Use the integral test to show that $\displaystyle\sum_{n=2}^\infty\frac{1}{n(\ln n)^2}$ converges, and bound its sum.
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Show that $\displaystyle\sum_{n=1}^\infty\frac{2 + \cos n}{n^2}$ converges, and give bounds for its sum.
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Does $\displaystyle\sum_{n=1}^\infty\frac{3n + 1}{n^3 + 2}$ converge?
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Show that $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n} = 1 - \frac12 + \frac13 - \dots$ converges, decide whether it converges absolutely, and bound the error of $S_4$
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How many terms of $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^3}$ guarantee an error less than $0.001$
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Use the ratio test on $\displaystyle\sum_{n=1}^\infty\frac{n!}{n^n}$
···
Proofs that stochastic gradient descent converges usually ask for step sizes $\alpha_n > 0$ with $\sum\alpha_n = \infty$ and $\sum\alpha_n^2 < \infty$ For which $p$ does $\alpha_n = \dfrac{c}{n^p}$ (with $c > 0$ qualify? Does $\alpha_n = 0.1\cdot0.99^n$

Answers

$10$
$1$
The first diverges; for the second the test is silent, and it diverges anyway
It converges, with $1.443 \le S \le 2.483$
It converges, to a sum between $1.645$ and $4.935$
Yes, it converges
It converges conditionally, not absolutely; $\lvert S - S_4\rvert \le 0.2$
$10$ terms
$\ell = \tfrac1e \approx 0.368 < 1$ so the series converges
$\tfrac12 < p \le 1$ the geometric schedule fails, because $\sum\alpha_n = 9.9$ is finite

Worked solutions

Problem 1

Find $\displaystyle\sum_{n=1}^\infty 5\Big(\frac23\Big)^n$

$r = \tfrac23$ and $\lvert r\rvert < 1$ so the series convergesA geometric series: each term is $\tfrac23$ of the one before.
The first term, at $n = 1$ is $5\cdot\tfrac23 = \tfrac{10}3$ The sum starts at $n = 1$ not $n = 0$
$\dfrac{10/3}{1 - 2/3} = \dfrac{10/3}{1/3}$ First term over $1 - r$
$10$ The partial sums $\tfrac{10}3$ $\tfrac{50}9$ $\tfrac{190}{27}, \dots$ climb toward $10$

Problem 2

Find $\displaystyle\sum_{n=1}^\infty \frac{1}{n(n + 1)}$ from its partial sums.

$\dfrac{1}{n(n + 1)} = \dfrac1n - \dfrac1{n + 1}$ Partial fractions.
$S_N = \big(1 - \tfrac12\big) + \big(\tfrac12 - \tfrac13\big) + \dots + \big(\tfrac1N - \tfrac1{N + 1}\big) = 1 - \dfrac1{N + 1}$ Every middle term cancels with the next one: the sum telescopes.
$1 - \dfrac1{N + 1} \to 1$ Take the limit of the partial sums.
$1$ For example $S_{99} = 0.99$ Telescoping sums are one of the few kinds, with geometric series, whose exact value is easy to find.

Problem 3

What does the divergence test say about $\displaystyle\sum_{n=1}^\infty\frac{n}{2n + 1}$ and about $\displaystyle\sum_{n=1}^\infty\frac1{\sqrt n}$

$\dfrac{n}{2n + 1} \to \dfrac12 \ne 0$ Divide top and bottom by $n$
So $\displaystyle\sum\frac{n}{2n + 1}$ divergesAdding terms near $\tfrac12$ forever gives partial sums that grow like $\tfrac N2$
$\dfrac1{\sqrt n} \to 0$ so the divergence test says nothing about $\sum\frac1{\sqrt n}$ Terms going to $0$ are needed for convergence, but they are not enough.
The first diverges; for the second the test is silent, and it diverges anywayIt is a $p$ -series with $p = \tfrac12 \le 1$ Directly: each of the first $N$ terms is at least $\tfrac1{\sqrt N}$ so $S_N \ge \sqrt N$ $S_{100} \approx 18.59$

Problem 4

Use the integral test to show that $\displaystyle\sum_{n=2}^\infty\frac{1}{n(\ln n)^2}$ converges, and bound its sum.

$f(x) = \dfrac{1}{x(\ln x)^2}$ is positive and decreasing for $x \ge 2$ Both $x$ and $\ln x$ increase, so the denominator does.
$\displaystyle\int_2^b\frac{dx}{x(\ln x)^2} = \Big[-\frac1{\ln x}\Big]_2^b = \frac1{\ln2} - \frac1{\ln b}$ Substitute $u = \ln x$ $du = \tfrac{dx}x$ the integral of $u^{-2}$
$\dfrac1{\ln b} \to 0$ so the integral is $\dfrac1{\ln2} \approx 1.4427$ and the series convergesThe integral test.
$\dfrac1{\ln2} \le S \le \dfrac{1}{2(\ln2)^2} + \dfrac1{\ln2}$ The bounds with $N = 2$ the integral alone, and the integral plus the first term.
It converges, with $1.443 \le S \le 2.483$ The sum is about $2.110$ Without the square, $\sum\frac{1}{n\ln n}$ diverges: its integral is $\ln(\ln b) \to \infty$

Problem 5

Show that $\displaystyle\sum_{n=1}^\infty\frac{2 + \cos n}{n^2}$ converges, and give bounds for its sum.

$1 \le 2 + \cos n \le 3$ so $\dfrac1{n^2} \le \dfrac{2 + \cos n}{n^2} \le \dfrac3{n^2}$ $\cos n$ is between $-1$ and $1$ Every term is positive.
$\displaystyle\sum\frac3{n^2}$ convergesA $p$ -series with $p = 2 > 1$ times $3$
So the series converges by direct comparisonSmaller than a convergent series, term by term.
$\displaystyle\sum\frac1{n^2} = \frac{\pi^2}6 \approx 1.645$ and $\displaystyle\sum\frac3{n^2} = \frac{\pi^2}2 \approx 4.935$ The known value of $\sum\frac1{n^2}$
It converges, to a sum between $1.645$ and $4.935$ The sum is about $3.614$ The integral test can't be used here, because the terms are not decreasing.

Problem 6

Does $\displaystyle\sum_{n=1}^\infty\frac{3n + 1}{n^3 + 2}$ converge?

For large $n$ $\dfrac{3n + 1}{n^3 + 2}$ behaves like $\dfrac{3n}{n^3} = \dfrac3{n^2}$ Keep the highest power on top and bottom.
$\dfrac{(3n + 1)/(n^3 + 2)}{1/n^2} = \dfrac{3n^3 + n^2}{n^3 + 2} \to 3$ Limit comparison with $b_n = \tfrac1{n^2}$
$0 < 3 < \infty$ and $\sum\frac1{n^2}$ convergesBoth series do the same thing.
Yes, it convergesA direct comparison works too: $3n + 1 \le 4n$ and $n^3 + 2 > n^3$ so the terms are below $\tfrac4{n^2}$ The sum is about $3.260$

Problem 7

Show that $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n} = 1 - \frac12 + \frac13 - \dots$ converges, decide whether it converges absolutely, and bound the error of $S_4$

$b_n = \tfrac1n$ decreases and tends to $0$ The conditions of the alternating series test.
So the series convergesEach partial sum overshoots the last in the other direction, by less each time.
$\displaystyle\sum\Big\lvert\frac{(-1)^{n+1}}{n}\Big\rvert = \sum\frac1n$ divergesThe harmonic series, a $p$ -series with $p = 1$
$S_4 = 1 - \tfrac12 + \tfrac13 - \tfrac14 = \tfrac7{12} \approx 0.5833$ with error at most $b_5 = \tfrac15$ The first omitted term.
It converges conditionally, not absolutely; $\lvert S - S_4\rvert \le 0.2$ The sum is $\ln2 \approx 0.6931$ so the actual error is about $0.110$ inside the bound.

Problem 8

How many terms of $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^3}$ guarantee an error less than $0.001$

$b_n = \tfrac1{n^3}$ decreases to $0$ so the error after $N$ terms is at most $\dfrac1{(N + 1)^3}$ The alternating series error bound.
$\dfrac1{(N + 1)^3} < 0.001$ needs $(N + 1)^3 > 1000$ so $N + 1 > 10$ $10^3 = 1000$ exactly, which is not enough for a strict inequality.
$N = 10$ the bound is $\tfrac1{1331} \approx 0.00075$ $N = 9$ gives a bound of exactly $0.001$
$10$ terms $S_{10} \approx 0.90112$ and the sum is $\tfrac34\zeta(3) \approx 0.90154$ the actual error is about $0.00043$

Problem 9

Use the ratio test on $\displaystyle\sum_{n=1}^\infty\frac{n!}{n^n}$

$\dfrac{a_{n+1}}{a_n} = \dfrac{(n + 1)!}{(n + 1)^{n+1}}\cdot\dfrac{n^n}{n!}$ Write out the ratio of consecutive terms.
$= \dfrac{(n + 1)\,n^n}{(n + 1)^{n+1}} = \dfrac{n^n}{(n + 1)^n} = \Big(\dfrac{n}{n + 1}\Big)^n$ $(n + 1)!/n! = n + 1$ which cancels one factor of $n + 1$
$\Big(\dfrac{n}{n + 1}\Big)^n = \dfrac{1}{(1 + 1/n)^n} \to \dfrac1e$ The limit that defines $e$
$\ell = \tfrac1e \approx 0.368 < 1$ so the series convergesThe sum is about $1.880$ Factorials and powers in the same term are what the ratio test is for: the comparison tests have nothing simple to compare with.

Problem 10

Proofs that stochastic gradient descent converges usually ask for step sizes $\alpha_n > 0$ with $\sum\alpha_n = \infty$ and $\sum\alpha_n^2 < \infty$ For which $p$ does $\alpha_n = \dfrac{c}{n^p}$ (with $c > 0$ qualify? Does $\alpha_n = 0.1\cdot0.99^n$

$\displaystyle\sum\frac{c}{n^p} = \infty$ exactly when $p \le 1$ The $p$ -series test.
$\displaystyle\sum\Big(\frac{c}{n^p}\Big)^2 = c^2\sum\frac1{n^{2p}} < \infty$ exactly when $2p > 1$ Again the $p$ -series test, now with exponent $2p$
Both hold when $\tfrac12 < p \le 1$ So $\tfrac cn$ qualifies, and $\tfrac{c}{\sqrt n}$ (whose squares form the harmonic series) and $\tfrac{c}{n^2}$ (whose sum is finite) do not.
$\displaystyle\sum_{n=1}^\infty 0.1\cdot0.99^n = \frac{0.1\cdot0.99}{1 - 0.99} = 9.9$ Geometric, with first term $0.099$
$\tfrac12 < p \le 1$ the geometric schedule fails, because $\sum\alpha_n = 9.9$ is finiteThe first condition lets the steps reach a minimum however far away it starts; with steps adding to $9.9$ the iterates can only travel $9.9$ times the largest gradient. The second condition makes the noise in the gradients average out.

Where this goes wrong

1. Starting the geometric sum at the wrong term

$\dfrac{a}{1 - r}$ is memorized with $a$ as the number in front of $r^n$

$\displaystyle\sum_{n=1}^\infty 5\Big(\frac23\Big)^n$ geometric with $r = \tfrac23$ Right so far: Problem 1, step 1.
“ $a = 5$ ”The habit that causes the mistake: $5$ is the term at $n = 0$ and this sum starts at $n = 1$
$\dfrac{5}{1 - 2/3} = 15$ That includes a term of $5$ that is not in the series. The formula needs the first term actually in the sum, $5\cdot\tfrac23 = \tfrac{10}3$ and the sum is $10$

2. Reading terms that go to 0 as convergence

The divergence test gets used backwards.

$\dfrac1{\sqrt n} \to 0$ Right so far: Problem 3, step 3.
“The terms go to $0$ so the series converges.”The habit that causes the mistake: the divergence test turned around.
$\displaystyle\sum\frac1{\sqrt n}$ convergesIt diverges: $S_N \ge \sqrt N$ and $S_{10000} \approx 198.5$ Terms that go to $0$ are necessary for convergence, not sufficient, and the harmonic series $\sum\frac1n$ is the standard example.

3. Comparing with a bigger divergent series

Any true inequality looks like a comparison.

$a_n = \dfrac{3n + 1}{n^3 + 2} > 0$ Right so far: Problem 6.
“ $a_n \le \tfrac4n$ for every $n \ge 1$ and $\sum\frac4n$ diverges.”The inequality is true: $(3n + 1)n \le 4(n^3 + 2)$ for every $n \ge 1$
So $\sum a_n$ divergesBeing smaller than a divergent series proves nothing; a series of zeros is smaller too. To prove divergence, $a_n$ must be at least the terms of a divergent series; to prove convergence, at most those of a convergent one. Here $a_n \le \tfrac4{n^2}$ and the series converges.

4. Using the alternating series test on terms that don't shrink to 0

The signs alternate, so the test seems to apply.

$\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac{n}{n + 1}$ with $b_n = \dfrac{n}{n + 1}$ The setup of Problem 7, on a different series.
“It alternates, so it converges.”The habit that causes the mistake: the test has two conditions, and only the alternation was checked.
The series converges $b_n \to 1$ not $0$ so the terms do not go to $0$ and the series diverges by the divergence test: its partial sums keep jumping by nearly $1$ The alternating series test needs $b_n$ to decrease to $0$ .

5. Taking a ratio limit of 1 as an answer

The ratio test is applied to every series, and its answer is read even at the boundary.

For $\sum\frac1n$ $\dfrac{a_{n+1}}{a_n} = \dfrac{n}{n + 1} \to 1$ Right so far: the same step as Problem 9, step 1.
“ $\ell \le 1$ so it converges.”The habit that causes the mistake: the boundary case lumped in with $\ell < 1$
$\sum\frac1n$ convergesIt diverges. At $\ell = 1$ the ratio test gives no answer: $\sum\frac1{n^2}$ also has ratio $\big(\tfrac{n}{n + 1}\big)^2 \to 1$ and converges. Any $p$ -series has ratio limit $1$ so for these, use the integral test or a comparison instead.

Print this set: series-convergence-tests.pdf (problems, answers, and worked solutions on separate pages).

Read the guide: Series convergence tests: which test to use, and what each one can't tell you