Alternating Series and Absolute Convergence

Q: What is the alternating series test?

If bₙ > 0, the bₙ are decreasing, and bₙ → 0, then the sum of (−1)ⁿbₙ converges.

Quick answer

An alternating series switches sign from term to term. If the sizes of its terms shrink to 0, it converges: the partial sums step back and forth across the sum, each step shorter than the last. Stopping after n terms leaves an error no bigger than the first term left out. A series converges absolutely if the series of absolute values converges, and absolute convergence implies convergence. The alternating harmonic series converges but not absolutely, which is called conditional convergence.

What you'll learn

Apply the alternating series test
Bound the error of a partial sum of an alternating series
Classify a series as absolutely convergent, conditionally convergent or divergent
Use absolute convergence for series with irregular signs

Series whose signs alternate

The harmonic series $1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots$ diverges. Flip every other sign and something changes:

1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots

An alternating series has the form $\sum (-1)^n b_n$ or $\sum (-1)^{n+1} b_n$ with every $b_n > 0$ .

The alternating series test

If $b_n > 0$ , the $b_n$ are decreasing, and $\lim_{n \to \infty} b_n = 0$ , then $\sum (-1)^{n+1} b_n$ converges.

Why the partial sums close in

Follow the partial sums of $1 - \tfrac{1}{2} + \tfrac{1}{3} - \cdots$ . Each step goes the opposite way from the last, and each is shorter. So the partial sums zigzag, and every zigzag is narrower than the one before.

the sum, ln 2

Partial sums of 1 − 1/2 + 1/3 − 1/4 + ⋯

The odd partial sums come down, the even ones go up, and each odd one stays above every even one. With the step lengths $b_n$ shrinking to $0$ , the two sides squeeze together on a single number. An alternating series converges when its steps shrink to zero, because each step undoes part of the last. That number, here, is $\ln 2$ .

The error bound

The squeeze also measures accuracy. The sum always lies between two consecutive partial sums, so stopping after $n$ terms leaves an error smaller than the next step:

\lvert S - S_n \rvert \le b_{n+1}

The sign of the error is known too: if the last term used was positive, $S_n$ is too big.

Absolute and conditional convergence

For a series with terms of either sign, look at the sizes alone. If $\sum \lvert a_n \rvert$ converges, the series converges absolutely, and an absolutely convergent series always converges: the signs can only cancel, not add. A series that converges while $\sum \lvert a_n \rvert$ diverges converges conditionally. It converges only because of the cancellation.

Series	Absolute values	Verdict
$\sum \tfrac{(-1)^n}{n^2}$	$\sum \tfrac{1}{n^2}$ converges	converges absolutely
$\sum \tfrac{(-1)^n}{n}$	$\sum \tfrac{1}{n}$ diverges	converges conditionally
$\sum (-1)^n \tfrac{n}{n + 1}$	terms do not approach $0$	diverges

Worked examples

Common mistakes

Practice problems

Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n + 1}$ .

Answer

Converges conditionally

Full solution

$\tfrac{1}{n + 1}$ decreases to $0$ , so the series converges. The absolute values form a shifted harmonic series, which diverges.
Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$ .

Answer

Converges absolutely

Full solution

$\sum \tfrac{1}{n^3}$ converges, a $p$ -series with $p = 3$ .
Classify $\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{n}{2n + 1}$ .

Answer

Diverges

Full solution

The terms do not approach $0$ : their sizes approach $\tfrac{1}{2}$ .
Classify $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n}$ .

Answer

Converges conditionally

Full solution

$\tfrac{1}{\ln n}$ decreases to $0$ , so the series converges. Since $\ln n < n$ , $\tfrac{1}{\ln n} > \tfrac{1}{n}$ , so the absolute values diverge by comparison with the harmonic series.
Estimate $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3}$ with three terms and bound the error.

Answer

About $0.912$ , within $0.016$

Full solution

$1 - \tfrac{1}{8} + \tfrac{1}{27} \approx 0.912$ , and the error is at most $\tfrac{1}{4^3} = \tfrac{1}{64}$ .
The series $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$ equals $\tfrac{1}{e}$ . How many terms guarantee an error below $0.001$ ?

Answer

$7$ terms, $n = 0$ to $6$

Full solution

Stopping after the $n = k$ term leaves an error at most $\tfrac{1}{(k + 1)!}$ . $\tfrac{1}{6!} = \tfrac{1}{720} > 0.001$ but $\tfrac{1}{7!} = \tfrac{1}{5040} < 0.001$ , so stop after $n = 6$ .
Classify $\displaystyle\sum_{n=1}^{\infty} (-1)^n \frac{n + 1}{n^2}$ .

Answer

Converges conditionally

Full solution

$\tfrac{n + 1}{n^2} = \tfrac{1}{n} + \tfrac{1}{n^2}$ decreases to $0$ , so the series converges. The absolute values are larger than $\tfrac{1}{n}$ , so they diverge.
Classify $\displaystyle\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{n}$ .

Answer

Converges conditionally

Full solution

$\cos(n\pi) = (-1)^n$ , so this is $\sum \tfrac{(-1)^n}{n}$ , the alternating harmonic series with its signs flipped.
Classify $\displaystyle\sum_{n=1}^{\infty} \frac{(-2)^n}{n^2}$ .

Answer

Diverges

Full solution

$\tfrac{2^n}{n^2}$ grows without bound, so the terms do not approach $0$ .
A student says $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ converges absolutely, because its terms approach $0$ . What went wrong?

Hint

What series do the absolute values form?

Answer

The absolute values form the harmonic series, which diverges. The series converges only conditionally.

Full solution

Terms approaching $0$ is a condition for convergence of any kind, not a proof of absolute convergence. Absolute convergence means $\sum \lvert a_n \rvert = \sum \tfrac{1}{n}$ converges, and it does not.

The series does converge, by the alternating series test, so it converges conditionally.

Frequently asked questions

What is the alternating series test?

If bₙ > 0, the bₙ are decreasing, and bₙ → 0, then the sum of (−1)ⁿbₙ converges.

How accurate is a partial sum of an alternating series?

The error after n terms is at most the size of the first omitted term, b₍ₙ₊₁₎, and the sum lies between any two consecutive partial sums.

What is absolute convergence?

A series converges absolutely if the series of absolute values of its terms converges. Absolutely convergent series always converge.

What is conditional convergence?

A series that converges but whose absolute values form a divergent series, such as 1 − 1/2 + 1/3 − 1/4 + ⋯.

Does the sum of an alternating series lie above or below a partial sum?

Between two consecutive partial sums. If the last term used was positive, the partial sum is too big; if negative, too small.

What to learn next

The Ratio and Root TestsTest series with factorials and powers by comparing each term with the one before: the ratio test, the root test, and what an inconclusive result means.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Series whose signs alternate

Why the partial sums close in

The error bound

Absolute and conditional convergence

Worked examples

Common mistakes

Practice problems

Frequently asked questions

What is the alternating series test?

How accurate is a partial sum of an alternating series?

What is absolute convergence?

What is conditional convergence?

Does the sum of an alternating series lie above or below a partial sum?

What to learn next

More Calculus lessons