10.7 - Alternating Series Test for Convergence

Definition of an Alternating Series

An alternating series is one whose terms flip sign: positive, negative, positive, negative, and so on.

\sum_{n=1}^{\infty} (-1)^n a_n \quad \text{or} \quad \sum_{n=1}^{\infty} (-1)^{n+1} a_n

where

a_n > 0

for all

n

The difference between the two: -

(-1)^n

: the first term is negative (when

n = 1

) -

(-1)^{n+1}

: the first term is positive (when

n = 1

)

Structurally, the

(-1)^n

(-1)^{n+1}

creates the sign pattern, and

a_n

is the positive magnitude of each term.

The Alternating Series Test (AST)

For

\sum_{n=1}^{\infty} (-1)^n a_n

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

, where

a_n > 0

The series CONVERGES if both conditions hold.

Condition 1: the terms approach zero

\lim_{n \to \infty} a_n = 0

Condition 2: the terms are decreasing

a_{n+1} \leq a_n \text{ for all } n

Equivalently,

a_1 \geq a_2 \geq a_3 \geq \cdots

\boxed{\text{If BOTH conditions hold, the alternating series CONVERGES}}

If either condition fails, the test is inconclusive. (And if Condition 1 fails, the series actually diverges by the nth term test.)

Checking if the Terms Are Decreasing

Three ways:

Method 1: take the derivative (least recommended). Treat

a_n

as a function

f(x)

and show

f'(x) < 0

past some point. It works, but it’s usually tedious.

Method 2: plug in values (most intuitive). Compute

a_1, a_2, a_3, a_4, \ldots

and see that each term is smaller than the last. Usually it’s obvious.

Method 3: algebraic manipulation (most rigorous). Show that

a_n - a_{n+1} > 0

. Or, often easier, show that:

\frac{a_{n+1}}{a_n} < 1

If the ratio is less than 1, then

a_{n+1} < a_n

Example 1: The Alternating Harmonic Series

Does

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

converge or diverge?

Step 1: identify it as alternating. It has the form

(-1)^{n+1} a_n

with

a_n = \frac{1}{n}

Written out:

1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots

Step 2: check Condition 1

\lim_{n \to \infty} \frac{1}{n} = 0 \quad ✓

Step 3: check Condition 2. The terms are

1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots

, and clearly:

1 > \frac{1}{2} > \frac{1}{3} > \frac{1}{4} > \cdots \quad ✓

Both conditions hold, so the series converges by the Alternating Series Test.

Worth pausing on: the regular harmonic series

\sum \frac{1}{n}

diverges, but the alternating harmonic series converges. That’s how much of a difference alternation makes.

Example 2: Rational Function with Alternating Sign

Does

\sum_{n=1}^{\infty} (-1)^{n} \frac{n+3}{n^2 + 4n + 5}

converge or diverge?

Step 1: identify $a_n$

a_n = \frac{n+3}{n^2 + 4n + 5}

Step 2: check Condition 1. The denominator has degree 2 and the numerator degree 1, so:

\lim_{n \to \infty} \frac{n+3}{n^2 + 4n + 5} = 0 \quad ✓

Step 3: check Condition 2 using the ratio method. We need

\frac{a_{n+1}}{a_n} < 1

a_{n+1} = \frac{(n+1)+3}{(n+1)^2 + 4(n+1) + 5} = \frac{n+4}{n^2 + 6n + 10}

Form the ratio:

\frac{a_{n+1}}{a_n} = \frac{\frac{n+4}{n^2 + 6n + 10}}{\frac{n+3}{n^2 + 4n + 5}} = \frac{(n+4)(n^2 + 4n + 5)}{(n+3)(n^2 + 6n + 10)}

Expand the numerator:

(n+4)(n^2 + 4n + 5) = n^3 + 8n^2 + 21n + 20

Expand the denominator:

(n+3)(n^2 + 6n + 10) = n^3 + 9n^2 + 28n + 30

Now compare. The denominator is exactly the numerator plus

(n^2 + 7n + 10)

, which is positive for all

n \geq 1

n^3 + 9n^2 + 28n + 30 = (n^3 + 8n^2 + 21n + 20) + (n^2 + 7n + 10)

Since the denominator is bigger than the numerator:

\frac{a_{n+1}}{a_n} < 1 \quad ✓

Both conditions hold, so the series converges by AST.

Example 3: When Condition 1 Fails

Does

\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n+5}{n}

converge or diverge?

Step 1: identify $a_n$

a_n = \frac{n+5}{n}

Step 2: check Condition 1

\lim_{n \to \infty} \frac{n+5}{n} = \lim_{n \to \infty} \left(1 + \frac{5}{n}\right) = 1

Since the limit is

1 \neq 0

, Condition 1 fails.

The series diverges.

You can stop right here. By the nth term test for divergence, if

\lim_{n \to \infty} a_n \neq 0

, the series has to diverge, alternating or not. No need to check whether the terms are decreasing.

Alternating Series Without an Explicit $(-1)^n$

Sometimes the alternation is hiding in a trig function.

Example 4: cosine creating alternation

Does

\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{n}

converge or diverge?

Step 1: recognize the pattern. Evaluate

\cos(n\pi)

\cos(\pi) = -1, \quad \cos(2\pi) = 1, \quad \cos(3\pi) = -1, \quad \cos(4\pi) = 1

\cos(n\pi) = (-1)^n

, which means:

\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}

That’s the alternating harmonic series in disguise.

Step 2: apply AST with $a_n = \frac{1}{n}$

Condition 1: $\lim_{n \to \infty} \frac{1}{n} = 0$ ✓

Condition 2: $\frac{1}{n}$ is decreasing ✓

The series converges by AST.

When a Series Is NOT Alternating

Careful. Not every series with a

(-1)^n

in it actually alternates.

Example 5: double alternation cancels out

\sum_{n=1}^{\infty} (-1)^n \cos(n\pi) \cdot \frac{1}{n^2 + 2n}

alternating?

Step 1: analyze the sign pattern. There are two alternating factors: -

(-1)^n

: alternates

-1, 1, -1, 1, \ldots

\cos(n\pi) = (-1)^n

: also alternates

-1, 1, -1, 1, \ldots

Step 2: multiply them

$n = 1$ : $(-1)^1 \cdot \cos(\pi) = (-1)(-1) = 1$ (positive)

$n = 2$ : $(-1)^2 \cdot \cos(2\pi) = (1)(1) = 1$ (positive)

$n = 3$ : $(-1)^3 \cdot \cos(3\pi) = (-1)(-1) = 1$ (positive)

The product is always positive.

This is not an alternating series. The two alternating factors cancel each other out, leaving every term positive. You’d need a different test.

Strategy for Alternating Series Problems

Step 1: confirm it’s actually alternating. - Look for

(-1)^n

(-1)^{n+1}

- Or check whether a trig function is creating the alternation - Watch for double alternation that cancels out

Step 2: apply Condition 1. - Compute

\lim_{n \to \infty} a_n

- Limit isn’t 0 → stop, the series diverges by the nth term test - Limit is 0 → continue

Step 3: apply Condition 2. - Show the terms are decreasing - Easiest: plug in values, or show

\frac{a_{n+1}}{a_n} < 1

Step 4: conclude. - Both conditions met → converges by AST - Either fails → the test is inconclusive (or it diverges, if Condition 1 failed)

Alternating vs. Non-Alternating

Series	Type	Result
$\sum \frac{1}{n}$	Harmonic (not alternating)	Diverges
$\sum \frac{(-1)^{n+1}}{n}$	Alternating harmonic	Converges
$\sum \frac{1}{n^2}$	p-series, $p = 2$	Converges
$\sum \frac{(-1)^n}{n^2}$	Alternating p-series	Converges

Alternation can turn a divergent series into a convergent one. But it doesn’t always change anything.

Common Mistakes to Avoid

Forgetting to check Condition 1. Always verify $\lim_{n \to \infty} a_n = 0$ first

Including the $(-1)^n$ in $a_n$ . $a_n$ is the positive magnitude only. Wrong: $a_n = \frac{(-1)^n}{n}$ . Right: $a_n = \frac{1}{n}$

Checking the wrong thing for Condition 2. You need $a_n$ decreasing, not $(-1)^n a_n$

Not recognizing cancellation. Check whether multiple alternating factors cancel

Stopping at Condition 1. Both conditions are required

Visual Understanding

10.7 - Alternating Series Test for Convergence

The partial sums oscillate above and below the limit, with the swings shrinking each time, eventually closing in on the value.

Key Takeaways

Alternating series: terms flip sign, usually via $(-1)^n$ or $(-1)^{n+1}$

Two conditions for convergence:

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

$a_n$ is decreasing ( $a_{n+1} \leq a_n$ )

Both are required. If either fails, AST doesn’t guarantee convergence

Condition 1 failing means the series diverges by the nth term test

Three ways to check decreasing: derivative (hard), plugging in values (intuitive), or the ratio $\frac{a_{n+1}}{a_n} < 1$ (rigorous)

The alternating harmonic series converges, even though the regular harmonic series diverges

Alternation can change behavior: some divergent series converge once you alternate them

$(-1)^n$ vs. $(-1)^{n+1}$ : the only difference is the starting sign

Watch for hidden alternation: $\cos(n\pi) = (-1)^n$ , while $\sin(n\pi) = 0$ (not alternating at all)

Double alternation cancels: $(-1)^n \cdot \cos(n\pi)$ is always positive

$a_n$ is the magnitude. Don’t include the $(-1)^n$ when you identify it

Back Next

10.7 - Alternating Series Test for Convergence

Definition of an Alternating Series

The Alternating Series Test (AST)

Checking if the Terms Are Decreasing

Example 1: The Alternating Harmonic Series

Example 2: Rational Function with Alternating Sign

Example 3: When Condition 1 Fails

Alternating Series Without an Explicit (−1)n(-1)^n(−1)n

When a Series Is NOT Alternating

Strategy for Alternating Series Problems

Alternating vs. Non-Alternating

Common Mistakes to Avoid

Visual Understanding

Key Takeaways

Alternating Series Without an Explicit $(-1)^n$