Alternating Series, calculus lesson

Take a step forward, then a shorter step back, then a still shorter step forward, and so on. Each move overshoots and then partly undoes the last one, so you zigzag around a single point and close in on it. Series whose signs alternate behave like this walk, and they can converge even when the sizes of their terms would add up to infinity.

Key terms

Alternating series test

Definition An alternating series has the form

\sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + \cdots \quad \text{with } b_n > 0,

or the same with $(-1)^n$ . If

the sizes never increase, $b_{n+1} \le b_n$ for all $n \ge 1$ , and
the sizes tend to zero, $\lim_{n \to \infty} b_n = 0$ ,

then the series converges.

The even partial sums $s_2, s_4, s_6, \ldots$ increase, the odd partial sums $s_1, s_3, s_5, \ldots$ decrease, every even one lies below every odd one, and their difference $s_{2k+1} - s_{2k} = b_{2k+1}$ tends to $0$ . So both close in on the same number.

In plain words If the signs alternate and the step sizes shrink to zero without ever growing, the zigzag settles down.

Example For $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$ we have $b_n = \frac{1}{\sqrt{n}} > 0$ , the sequence $\frac{1}{\sqrt{n}}$ decreases, and it tends to $0$ . So the series converges, even though $\sum \frac{1}{\sqrt{n}}$ diverges.

Decreasing magnitudes that tend to zero

Definition The decreasing condition may be weakened to $b_{n+1} \le b_n$ for all $n \ge N$ , for some integer $N$ , because finitely many terms do not affect convergence. The condition $b_n \to 0$ cannot be dropped, and neither can the decreasing condition. A common way to check decreasing is to write $b_n = f(n)$ for a function $f$ that is differentiable on $[N, \infty)$ and show $f'(x) \le 0$ for $x \ge N$ .

In plain words The sizes only need to go downhill from some point on. They must go to zero, and they must not jump back up along the way.

Example For $\sum_{n=2}^{\infty} (-1)^n \frac{\ln n}{n}$ take $f(x) = \frac{\ln x}{x}$ . Its derivative is

f'(x) = \frac{1 - \ln x}{x^2} < 0 \quad \text{for } x > e,

so $b_n$ decreases for $n \ge 3$ . Also $\frac{\ln n}{n} \to 0$ , so the series converges.

Compare $\sum (-1)^n \frac{n}{n + 1}$ . Here $b_n = \frac{n}{n + 1} \to 1$ , so the terms do not go to zero and the series diverges by the nth term test.

Decreasing matters too. The series $1 - 1 + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{9} + \frac{1}{4} - \frac{1}{16} + \cdots$ alternates the positive terms $\frac{1}{k}$ with the negative terms $-\frac{1}{k^2}$ . Its sizes tend to $0$ but do not decrease, since $\frac{1}{4}$ is followed by $\frac{1}{3}$ . It diverges, because its partial sums with an even number of terms are

s_{2K} = H_K - \sum_{k=1}^{K} \frac{1}{k^2} \ge H_K - 2 \to \infty,

where $H_K$ is the $K$ th harmonic partial sum and every partial sum of $\sum \frac{1}{k^2}$ is below $2$ .

Alternating harmonic series

Definition The alternating harmonic series is

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

It converges by the alternating series test, since $\frac{1}{n}$ is positive, decreasing and tends to $0$ . Its sum is $\ln 2$ , a fact that follows from the Taylor series of $\ln(1 + x)$ .

In plain words The harmonic series diverges, but flipping every other sign makes it converge. The positive and negative parts nearly cancel.

Example The partial sums zigzag around $\ln 2 \approx 0.6931$ .

$N$	$1$	$2$	$3$	$4$	$5$	$6$
$s_N$	$1$	$0.5$	$0.8333$	$0.5833$	$0.7833$	$0.6167$

Odd partial sums sit above $\ln 2$ and even ones below.

Alternating remainder estimate

Definition Suppose $\sum (-1)^{n-1} b_n$ satisfies $b_n > 0$ , $b_{n+1} \le b_n$ for all $n \ge 1$ , and $b_n \to 0$ , with sum $S$ . Then for every $n \ge 1$

\lvert R_n \rvert = \lvert S - s_n \rvert \le b_{n+1}.

Moreover $S - s_n$ has the same sign as the first omitted term $(-1)^n b_{n+1}$ , or is zero.

In plain words When you stop, the error is no bigger than the size of the next term you would have added, and it points the same way as that term.

Example How many terms of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^3}$ make the error less than $0.001$ ? The sizes $\frac{1}{n^3}$ are positive, decreasing and tend to $0$ , so it is enough that

b_{n+1} = \frac{1}{(n + 1)^3} < 0.001,

that is $(n + 1)^3 > 1000$ . So $n + 1 \ge 11$ and $n = 10$ terms suffice. Indeed $s_{10} \approx 0.901116$ , and the true sum is $\frac{3}{4}\zeta(3) \approx 0.901543$ , an error of about $0.00043$ , below $b_{11} = \frac{1}{1331} \approx 0.00075$ . Since the first omitted term is positive, $s_{10}$ is below the sum, as it should be.

Check yourself

Does $\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n}$ converge?

Show the answer

Yes. The sizes $\frac{1}{\ln n}$ are positive, decrease because $\ln n$ increases, and tend to $0$ . The alternating series test applies.

Does $\sum_{n=1}^{\infty} (-1)^n \frac{n^2}{n^2 + 1}$ converge?

Show the answer

No. The sizes tend to $1$ , so the terms do not tend to $0$ and the series diverges.

Estimate $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!}$ with error less than $0.01$ .

Show the answer

The sizes $\frac{1}{n!}$ are positive and decrease to $0$ . We need $\frac{1}{(n + 1)!} < 0.01$ , so $(n + 1)! > 100$ , which first holds at $n + 1 = 5$ . Then

s_4 = 1 - \frac{1}{2} + \frac{1}{6} - \frac{1}{24} = 0.625,

within $\frac{1}{120} \approx 0.0083$ of the sum. The exact value is $1 - \frac{1}{e} \approx 0.632$ .

Alternating Series

Key terms

Alternating series test

Decreasing magnitudes that tend to zero

Alternating harmonic series

Alternating remainder estimate

Check yourself

How this lesson connects

Builds on

Used later in