Unit 10 · Topic 10.7

The Alternating Series Test

A series whose terms alternate in sign converges when the sizes of its terms decrease and approach 0.

Practice 3: JustificationPractice 4: Communication and Notation

Part 1 · Hook

Why this matters

Add

1

, subtract

\frac{1}{2}

, add

\frac{1}{3}

, subtract

\frac{1}{4}

, and keep going. Without the signs the sum grows without bound, but with them each step undoes part of the last one, and the running total settles. The alternating series test says exactly when that happens.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What does the nth term test say about $\sum a_n$ when $\lim_{n \to \infty} a_n = 0$ ?

Nothing: the series may converge or diverge
The series converges
The series diverges
The series converges to 0

Show the answer

A limit of 0 leaves the question open; the nth term test can only show divergence.

Correct: Nothing: the series may converge or diverge:
The series converges:
The series diverges:
The series converges to 0:

2. Does the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ converge?

No: it diverges, though its terms approach 0
Yes: its terms approach 0
Yes: it is a $p$ -series with $p = 1$
Yes: its sum is 1

Show the answer

It is the $p$ -series with $p = 1$ , which diverges.

Correct: No: it diverges, though its terms approach 0:
Yes: its terms approach 0:
Yes: it is a $p$ -series with $p = 1$ :
Yes: its sum is 1:

3. Which sequence is decreasing for $n \ge 1$ ?

$\frac{1}{n + 2}$
$\frac{n}{n + 2}$
$(-1)^n$
$\sin n$

Show the answer

A larger $n$ gives a larger denominator, so $\frac{1}{n + 2}$ decreases.

Correct: $\frac{1}{n + 2}$ :
$\frac{n}{n + 2}$ :
$(-1)^n$ :
$\sin n$ :

Part 4 · See it

See it first

The partial sums of

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

zigzag in on the sum, about

0.693

. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The signs alternateeach partial sum steps up, then down, then up again, so the partial sums zigzag
The sizes of the terms decreaseeach step is shorter than the last, so the zigzag never passes back beyond the previous turn
The sizes approach 0the steps shrink to nothing, the zigzag closes in on one number, and the series converges

Part 6 · Key ideas

Key ideas

The alternating series test: if $a_n > 0$ , the $a_n$ decrease (from some point on), and $\lim_{n \to \infty} a_n = 0$ , then $\sum (-1)^n a_n$ converges.
Check and state all three conditions: alternating signs, decreasing sizes, a limit of 0.
The test only proves convergence. If the sizes do not approach 0, use the nth term test for divergence instead.
When the conditions hold from the first term on, the sum lies between any two consecutive partial sums.

Part 7 · Misconception

A common mistake

The wrong idea: The alternating series test fails for $\sum (-1)^n \frac{n}{n^2 + 3}$ because $a_2 > a_1$ , so the series diverges.

What actually happens: The sizes only need to decrease from some point on (here from $n = 2$ ), so the test applies and the series converges. And a failed condition never proves divergence by itself.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Partial sums of $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$

The points show the partial sums

S_1, S_2, \ldots, S_8

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

= 1 - \frac{1}{\sqrt{2}}

+ \frac{1}{\sqrt{3}} - \cdots

. The dashed line is at the sum

S

of the series.

1. Which statement verifies the conditions of the alternating series test for this series?

The terms alternate in sign and $\sum \frac{1}{\sqrt{n}}$ diverges
The terms alternate in sign and the partial sums stay between 0 and 1
The signs alternate and $\frac{1}{\sqrt{n}}$ decreases to 0
The sizes $\frac{1}{\sqrt{n}}$ decrease, so the partial sums decrease

Show the answer

The series is $\sum (-1)^{n+1} a_n$ with $a_n = \frac{1}{\sqrt{n}} > 0$ ; $a_{n+1}$ $= \frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}}$ $= a_n$ ; and $\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$ . So the series converges by the alternating series test.

The terms alternate in sign and $\sum \frac{1}{\sqrt{n}}$ diverges: What $\sum \frac{1}{\sqrt{n}}$ does is not a condition of this test; the sizes must decrease to 0.
The terms alternate in sign and the partial sums stay between 0 and 1: Bounded partial sums are not a condition of the test; the sizes must decrease and approach 0.
Correct: The signs alternate and $\frac{1}{\sqrt{n}}$ decreases to 0: Right: all three conditions, each one stated.
The sizes $\frac{1}{\sqrt{n}}$ decrease, so the partial sums decrease: The partial sums go up and down; the test needs alternating signs and a limit of 0 as well.

2. Which statement about the sum $S$ does the graph support?

$S_8 < S < S_7$
$S_7 < S < S_8$
$S = S_8$ , the last partial sum shown
$S > S_1 = 1$

Show the answer

The partial sums alternate above and below the sum while their swings shrink, so $S$ lies between any two consecutive partial sums. $S_7$ is above the line and $S_8$ below it: $S_8 < S < S_7$ .

Correct: $S_8 < S < S_7$ : Right: consecutive partial sums sit on either side of the sum.
$S_7 < S < S_8$ : $S_7$ ends with an added term, so it is above the sum, and $S_8$ is below it.
$S = S_8$ , the last partial sum shown: $S_8$ ends with a subtracted term and lies below the dashed line.
$S > S_1 = 1$ : $S_1 = 1$ is the highest point; every later partial sum and the sum lie below it.

Table

The sizes of the terms of an alternating series

$a_n = \frac{n}{n^2 + 3}$ for the series $\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2 + 3}$
$n$	$1$	$2$	$3$	$4$	$5$
$a_n$	$\frac{1}{4}$	$\frac{2}{7}$	$\frac{3}{12}$	$\frac{4}{19}$	$\frac{5}{28}$

3. What does the alternating series test conclude about $\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2 + 3}$ ?

Nothing, since $a_2 > a_1$ breaks the decreasing condition
It diverges, since $a_2 > a_1$
It converges, since the terms alternate in sign
It converges: the sizes decrease from $n = 2$ on and approach 0

Show the answer

The signs alternate; $a_n = \frac{n}{n^2 + 3}$ decreases for $n \ge 2$ ; and $\lim_{n \to \infty} \frac{n}{n^2 + 3}$ $= 0$ . Leaving out the first term does not change whether a series converges, so the series converges by the alternating series test.

Nothing, since $a_2 > a_1$ breaks the decreasing condition: The condition only has to hold from some point on; dropping the first term does not change convergence.
It diverges, since $a_2 > a_1$ : The test never shows divergence, and one early rise does not stop convergence.
It converges, since the terms alternate in sign: Alternating signs alone are not enough; the sizes must also decrease and approach 0.
Correct: It converges: the sizes decrease from $n = 2$ on and approach 0: Right: the first term does not affect convergence, so decreasing from $n = 2$ on is enough.

4. What happens to $\sum_{n=1}^{\infty} (-1)^n \frac{n + 1}{n}$ ?

It converges by the alternating series test
It diverges by the nth term test: the terms do not approach 0
It converges, since the sizes $\frac{n+1}{n}$ decrease
Nothing can be decided, since the alternating series test fails

Show the answer

$\frac{n + 1}{n} \to 1$ , so the terms $(-1)^n \frac{n+1}{n}$ do not approach 0 (they swing near $1$ and $-1$ ). By the nth term test, the series diverges.

It converges by the alternating series test: The sizes $\frac{n+1}{n}$ approach 1, not 0, so that test does not apply.
Correct: It diverges by the nth term test: the terms do not approach 0: Right: the sizes approach 1, so the terms swing near $\pm 1$ .
It converges, since the sizes $\frac{n+1}{n}$ decrease: Decreasing sizes are not enough; they must also approach 0.
Nothing can be decided, since the alternating series test fails: When the terms do not approach 0, the nth term test shows divergence.

5. For $\sum (-1)^n a_n$ with $a_n > 0$ , which conditions does the alternating series test need?

$a_{n+1} \le a_n$ from some $n$ on, and $\lim_{n \to \infty} a_n = 0$
$\lim_{n \to \infty} a_n = 0$ , with no condition on how the sizes change
$a_{n+1} \le a_n$ from some $n$ on, with no condition on the limit
$\sum a_n$ converges as a series of positive terms

Show the answer

The alternating series test: if the sizes $a_n$ are positive, decrease (at least from some point on), and approach 0, then $\sum (-1)^n a_n$ converges.

Correct: $a_{n+1} \le a_n$ from some $n$ on, and $\lim_{n \to \infty} a_n = 0$ : Right: decreasing sizes that approach 0.
$\lim_{n \to \infty} a_n = 0$ , with no condition on how the sizes change: A limit of 0 is not enough; the sizes must also decrease.
$a_{n+1} \le a_n$ from some $n$ on, with no condition on the limit: Decreasing is not enough: sizes that level off at 1 decrease but do not approach 0.
$\sum a_n$ converges as a series of positive terms: If $\sum a_n$ converges, the signed series does too, but this is not the alternating series test's condition.

6. Why does the alternating series test not apply to $\sum_{n=1}^{\infty} \frac{(-1)^{2n}}{n}$ ?

The sizes $\frac{1}{n}$ do not decrease as $n$ grows large
The sizes $\frac{1}{n}$ do not approach 0 as $n$ grows large
It does apply, and it shows that this series converges
$(-1)^{2n} = 1$ , so the terms are all positive and the series is harmonic

Show the answer

$(-1)^{2n}$ $= \left((-1)^2\right)^n$ $= 1^n$ $= 1$ , so the series is $\sum \frac{1}{n}$ , the harmonic series, which diverges. The alternating series test needs signs that alternate.

The sizes $\frac{1}{n}$ do not decrease as $n$ grows large: $\frac{1}{n}$ does decrease; the trouble is the sign.
The sizes $\frac{1}{n}$ do not approach 0 as $n$ grows large: $\frac{1}{n}$ approaches 0; the trouble is the sign.
It does apply, and it shows that this series converges: $(-1)^{2n} = \left((-1)^2\right)^n = 1$ : the series is $\sum \frac{1}{n}$ , which diverges.
Correct: $(-1)^{2n} = 1$ , so the terms are all positive and the series is harmonic: Right: an even power of $-1$ is 1, so there is no alternation.

Part 9 · Summary

Summary

A series whose terms alternate in sign converges when the sizes of its terms decrease and approach 0. Its partial sums zigzag around the sum with shrinking swings. If the sizes do not approach 0, the nth term test shows divergence; the alternating series test itself never shows divergence.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Partial sums of ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}

The sizes of the terms of an alternating series

Summary

What comes next

Connections

Partial sums of $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$