# The p-Series Test Examples 1

Recall from The p-Series Test page that:

Theorem (p-Series Test): The special series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ is convergent if $p > 1$ and divergent if $p ≤ 1$. |

We will now look at some examples of specifically applying the p-Series test.

## Example 1

**Using the p-Series test determine if the series $\sum_{n=1}^{\infty} \frac{n^2}{n^4}$ is convergent or divergent.**

Simplifying this series down we get that $\sum_{n=1}^{\infty} \frac{n^2}{n^4} = \sum_{n=1}^{\infty} \frac{1}{n^2}$ and so $p = 2$. Since $p = 2 > 1$, then by the p-series test this series is convergent.

## Example 2

**Using the p-Series test determine if the series $\sum_{n=1}^{\infty} \frac{\sec ^2 n - \tan ^2 n}{n}$ is convergent or divergent.**

We note the trigonometric identity that $\sec ^2 n - \tan ^2 n = 1$, and so $\sum_{n=1}^{\infty} \frac{\sec ^2 n - \tan ^2 n}{n} = \frac{1}{n}$, and so $p = 1$. Since $p = 1 ≤ 1$ we have that this series is divergent.

## Example 3

**Using the p-Series test determine if the series $\sum_{n=1}^{\infty} \frac{\cos (2\pi n)}{n^3}$ is convergent or divergent.**

We note that $\cos (2 \pi n) = 1$ for every $n \in \mathbb{N}$ and so $\sum_{n=1}^{\infty} \frac{\cos (2\pi n)}{n^3} = \frac{1}{n^3}$. So $p = 3$ and since $p = 3 ≥ 1$, then by the p-series test this series is convergent.