# The Alternating Series Test for Alternating Series of Real Numbers Examples 1

Recall from The Alternating Series Test for Alternating Series of Real Numbers page the following test for convergence of an alternating series:

The Alternating Series Test

If $(a_n)_{n=1}^{\infty}$ is a decreasing sequence of real numbers and $\displaystyle{\lim_{n \to \infty} a_n = 0}$, then we have the following conclusion:

- The alternating series $\displaystyle{\sum_{n=1}^{\infty} (-1)^{n+1} a_n}$ converges.

We will now look at some examples of applying the alternating series test.

## Example 1

**Show that $\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^n}{n + n^2}}$ converges.**

We first notice that $(a_n)_{n=1}^{\infty} = \left ( \frac{1}{n + n^2} \right )_{n=1}^{\infty}$ is a decreasing sequence of positive real numbers. To show this, let $n \in \mathbb{N}$. Then $n^2 < (n+1)^2$ Furthermore, $n + n^2 < (n+1) + (n+1)^2$ and so:

(1)Furthermore, it's not hard to see that:

(2)So, by the alternating series test, $\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^n}{n + n^2}}$ converges.

## Example 2

**Show that $\displaystyle{\sum_{n=1}^{\infty} \frac{5\cos(n \pi)}{2^n + 3^n}}$ converges.**

We note that the sequence $\displaystyle{(a_n)_{n=1}^{\infty} = \left ( \frac{5}{2^n + 3^n} \right )_{n=1}^{\infty}}$ is a decreasing sequence of real numbers. To show this, let $n \in \mathbb{N}$. Then $n < n+1$. So $2^n < 2^{n+1}$ and $3^n < 3^{n+1}$, so $2^n + 3^n < 2^{n+1} + 3^{n+1}$. Thus:

(3)Furthermore we have that:

(4)Also notice that $\cos(n\pi) = (-1)^{n}$ for all $n \in \mathbb{N}$.

So, the alternating series $\displaystyle{\sum_{n=1}^{\infty} \frac{5\cos(n \pi)}{2^n + 3^n}}$ converges by the alternating series test.