The Alternating Series Test

# The Alternating Series Test

So far we have looked at the following tests to determine if a series was convergent or divergent:

We haven't been able to use any of these tests to determine if a negative or partially negative series was convergent or divergent though. The following test will allow us to do so.

 Theorem (The Alternating Series Test): Let $\{ a_n \}$ be a sequence. If for $n$ sufficiently large, $a_na_{n+1} < 0$, $\mid a_{n+1} \mid ≤ \mid a_n \mid$, and $\lim_{n \to \infty} a_n = 0$, then the series $\sum_{n=1}^{\infty} a_n$ is convergent.

We note that the alternating series test has three requirements for $n$ sufficiently large. First, the terms must be alternating signs on consecutive terms. Secondly, the absolute value of terms must be decreasing in size. And lastly, the limit of the sequence of terms must approach 0. Under these conditions we can conclude that the series $\sum_{n=1}^{\infty} a_n$ is convergent.

• Proof of Theorem: Let $a_1 > 0$. Since $a_na_{n+1} < 0$ we get that $a_{2n+1} > 0$ and $a_{2n} < 0$ $\forall n \in \mathbb{N}$. Now let $s_n = a_1 + a_2 + ... + a_n$ denote the $n^{\mathrm{th}}$ partial sum of the series.
• Now since the terms are decreasing in size it follows that $a_{2n+1} ≥ -a_{2n+2}$ and so $s_{2n+2} = s_{2n} + a_{2n+1} + a_{2n+2} ≥ s_{2n}$. So the even partial sums $\{ s_{2n} \}$ form an increasing sequence.
• Similarly since the terms are decreasing in size it follows that $-a_{2n} ≥ a_{2n+1}$, and so $s_{2n+1} = s_{2n-1} + a_{2n} + a_{2n+1} ≤ s_{2n-1}$ and so the odd partial sums form a decreasing sequence $\{ s_{2n-1} \}$, and so:
(1)
\begin{equation} s_2 ≤ s_4 ≤ ... ≤ s_{2n} ≤ s_{2n-1} ≤ s_{2n-3} ≤ ... ≤ s_3 ≤ s_1 \end{equation}
• So $s_2$ is a lower bound for the sequence $\{ s_{n-1} \}$ and $s_1$ is an upper bound for the sequence $\{ s_{2n} \}$, both of which sequences converge, and so $\lim_{n \to \infty} s_{2n-1} = L_1$ and $\lim_{n \to \infty} s_{2n} = L_2$ by the monotonic sequence theorem.
• Now since we were given that $\lim_{n \to \infty} a_n = 0$ and we know that $a_{2n} = s_{2n} - s_{2n-1}$ then $0 = \lim_{n \to \infty} a_{2n} = \lim_{n \to \infty} s_{2n} - \lim_{n \to \infty} s_{2n-1} = L_1 - L_2$ and so $0 = L_1 - L_2$ which implies $L_1 = L_2$. So let $L = L_1 = L_2$ and so $\lim_{n \to \infty} s_n = L$ since every partial sum $s_n$ have been acknowledged, and therefore $\sum_{n=1}^{\infty} a_n$ is convergent to $L$.$\blacksquare$

We note that a similar proof works if the first term of the series is negative, that is $a_1 < 0$. We will now look at some examples applying the alternating series test.

## Example 1

Using the alternating series test determine if $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ is convergent or divergent.

We must first check to see if all of the conditions for the alternating series test are met before applying it. We note that $a_na_{n+1} < 0$ since the terms are alternating signs. We need to check if $\mid a_{n+1} \mid ≤ \mid a_n \mid$ for $n$ sufficiently large.

We note that $\mid a_{n+1} \mid = \biggr \rvert \frac{(-1)^{n+1}}{n+1} \biggr \rvert = \frac{1}{n+1}$ and that $\mid a_n \mid = \biggr \rvert \frac{(-1)^{n}}{n} \biggr \rvert = \frac{1}{n}$. We know that $\mid a_{n+1} \mid = \frac{1}{n+1} ≤ \frac{1}{n} = \mid a_n \mid$ so these terms are decreasing in size.

Lastly we note that $\lim_{n \to \infty} \frac{-1}{n} = 0$. So by the alternating series test, $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ is convergent.