The Alternating Series Test

An alternating series has terms that switch sign, one positive and the next negative. It can be written with a factor

(-1)^{n+1}

(-1)^n

in front of positive numbers

b_n

, as in

\begin{aligned} &\sum_{n=1}^{\infty} (-1)^{n+1} b_n \\[4pt] &\qquad = b_1 - b_2 + b_3 - b_4 + \cdots \end{aligned}

The test

Follow the partial sums of

b_1 - b_2 + b_3 - \cdots

when the

b_n

decrease. Each step reverses direction and is no longer than the step before. So the odd partial sums keep falling and the even ones keep rising, and every odd one stays above every even one. When the decrease starts only at some

N

, the same picture holds from there on.

The partial sums of the series in Example 1(a). Each one overshoots the sum, from alternate sides, by less than the one before.

When the steps shrink to 0, the two sequences squeeze together on a single number, the sum of the series. The test only ever proves convergence. If its conditions fail, it says nothing, and another test has to decide.

Example 1

Show that each series converges.

(a)

\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1}\big(e^{1/n} - 1\big)

(b)

\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1}\,\frac{n^2}{2^n}

(a) Here

b_n = e^{1/n} - 1

, which is positive since

\frac{1}{n} > 0

. As

n

increases,

\frac{1}{n}

decreases, and

e^x

is increasing, so

b_n

decreases. Since

\frac{1}{n} \to 0

b_n \to e^0 - 1 = 0

. Both conditions hold, so the series converges by the alternating series test.

(b) Here

b_n = \frac{n^2}{2^n} > 0

. Check where

f(x) = \frac{x^2}{2^x}

decreases.

f'(x) = \frac{x\,(2 - x\ln 2)}{2^x}

This is negative for

x > \frac{2}{\ln 2} \approx 2.89

, so

b_{n+1} \le b_n

for

n \ge 3

. The first few terms rise, from

b_1 = 0.5

b_3 = 1.125

. Dropping the first two terms doesn't change whether the series converges, so decreasing from

n = 3

on is enough. For the limit, apply L'Hospital's rule twice to

\frac{x^2}{2^x}

\begin{aligned} \lim_{x\to\infty} \frac{x^2}{2^x} &= \lim_{x\to\infty} \frac{2x}{2^x \ln 2} \\[4pt] &= \lim_{x\to\infty} \frac{2}{2^x (\ln 2)^2} = 0 \end{aligned}

Both conditions hold from

n = 3

on, so the series converges by the alternating series test.