The Alternating Series Test Examples 1

Let $\{ a_n \}$ be a sequence. Recall from The Alternating Series Test page that if for $n$ sufficiently large we have that:

• $a_na_{n+1} < 0$

• $\mid a_{n+1} \mid ≤ \mid a_n \mid$

• $\lim_{n \to \infty} a_n = 0$

Then the series $\sum_{n=1}^{\infty} a_n$ is convergent.

We will now look at some examples applying this test.

Example 1

Using the alternating series test, determine whether the series $\sum_{n=1}^{\infty} \frac{(-1)^n(n + 1)}{n^2}$.

We note that $a_na_{n+1} < 0$ since the terms in this series alternate due to $(-1)^n$. We also know that this series is ultimately decreasing since the denominator of the $n^{\mathrm{th}}$ term gets larger much faster than the numerator. Now let's check to see if $\lim_{n \to \infty} a_n = 0$.

(1)

\begin{align} \lim_{n \to \infty} \frac{(-1)^n(n + 1)}{n^2} = \lim_{n \to \infty} \frac{(-1)^n\left [ \frac{1}{n} + \frac{1}{n^2} \right ]}{1} = 0 \end{align}

Therefore by the alternating series test, $\sum_{n=1}^{\infty} \frac{(-1)^n(n + 1)}{n^2}$ is convergent.

Example 2

Using the alternating series test, determine whether the series $\sum_{n=1}^{\infty} \frac{\cos ( \pi n)n^4 + \cos (\pi n + 2 \pi)n}{n^5 - 2}$.

In this example, we note that $\cos (\pi n) = \cos (\pi n + 2\pi ) = (-1)^n$, and making this substitution we get:

(2)

\begin{align} \sum_{n=1}^{\infty} \frac{\cos ( \pi n)n^4 + \cos (\pi n + 2 \pi)n}{n^5 - 2} = \sum_{n=1}^{\infty} \frac{(-1)^n n^4 + (-1)^n n}{n^5 - 2} = \sum_{n=1}^{\infty} \frac{(-1)^n(n^4 + n)}{n^5 - 2} \end{align}

We note that this is an alternating series, that is $a_na_{n+1} < 0$. We also know that this series is ultimately decreasing in value. Evaluating the limit we get that:

(3)

\begin{align} \lim_{n \to \infty} \frac{(-1)^n(n^4 + n)}{n^5 - 2} = \lim_{n \to \infty} \frac{(-1)^n\left [ \frac{1}{n} + \frac{1}{n^4}\right ]}{1 - \frac{2}{n^5}} = 0 \end{align}

And so by the alternating series test, this series is convergent.