Alternating Series

An alternating series is an infinite series of the form:

$\displaystyle{\sum_{n=0}^\infty (-1)^n\,a_n}$

or:

$\displaystyle{\sum_{n=0}^\infty (-1)^{n-1}\,a_n}$

with $a_n > 0$ for all $n$. The signs of the general terms alternate between positive and negative. Like any series, an alternating series converges if and only if the associated sequence of partial sums converges.

Alternating Series Test

The theorem known as the "Leibniz Test," or the alternating series test, tells us that an alternating series will converge if the terms $a_n$ converge to $0$ monotonically.

Proof: Suppose the sequence $a_n$ converges to $0$ and is monotone decreasing. If $m$ is odd and $S_m - S_n < a_{m}$ via the following calculation: 

$\begin{aligned} S_m - S_n & = \sum_{k=0}^m(-1)^k\,a_k\,-\,\sum_{k=0}^n\,(-1)^k\,a_k\ \\& = \sum_{k=m+1}^n\,(-1)^k\,a_k \\ & =a_{m+1}-a_{m+2}+a_{m+3}-a_{m+4}+\cdots+a_n\\ & =\displaystyle a_{m+1}-(a_{m+2}-a_{m+3}) -\cdots-a_n \le a_{m+1}\le a_{m} \\& \quad [a_{n} \text{ decreasing}]. \end{aligned}$

Since $a_n$ is monotonically decreasing, the terms are negative. Thus, we have the final inequality $S_m - S_n \le a_{m}$. Similarly, it can be shown that, since $a_m$ converges to $0$, $S_m - S_n$ converges to $0$ for $m, n \rightarrow \infty$. Therefore, our partial sum $S_m$ converges. (The sequence $\{ S_m \}$ is said to form a Cauchy sequence, meaning that elements of the sequence become arbitrarily close to each other as the sequence progresses.) The argument for $m$ even is similar.

Example: 

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

$a_n = \frac1n$ converges to 0 monotonically. Therefore, the sum $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ converges by the alternating series test.

Alternating Harmonic Series

The first fourteen partial sums of the alternating harmonic series (black line segments) shown converging to the natural logarithm of 2 (red line).