# The p-Series Test

We will now look at another test for determining whether a series is convergent or divergent known as the **p-Series Test**. This test is actually a special case of the The Integral Test for Positive Series and is as follows:

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

**Proof of Theorem:**Consider the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$. Either this series converges or diverges. We note that $\frac{1}{n^p}$ is a positive decreasing series and the function $f(x) = \frac{1}{x^p}$ is continuous. Therefore we apply the integral test for positive series. First, consider the case where $p > 1$ and let's look at the following integral:

- Since $p > 1$, it follows that $\int \frac{1}{x^p} \: dx = \int x^{-p} \: dx = \frac{1}{(1 - p)x^{p - 1}}$, and so:

- Now since $p > 1$, the exponent $p - 1 > 0$ and thus:

- So if $p > 1$, the integral $\int_{1}^{\infty} \frac{1}{x^p} \: dx$ converges and so the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ must also be convergent by the integral test.

- Now suppose that $p = 1$. We will now evaluate the following integral:

- Therefore if $p = 1$, the series $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent by the integral test. (This should make sense as we just looked at The Harmonic Series).

- Lastly, suppose $p < 1$. Note that $\sum_{n=1}^{\infty} \frac{1}{n^p} \sum_{n=1}^{\infty} n^{-p}$ Since $-p > 1$ we have that by the divergence test that $\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges. $\blacksquare$

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

## Example 1

**Using the p-Series test, determine whether the series $\sum_{n=1}^{\infty} \frac{1}{n^5}$ is convergent or divergent.**

By the p-Series test we note that $p = 5 > 1$ and therefore the series is convergent.

## Example 2

**Using the p-Series test, determine whether the series $\sum_{n = 1}^{\infty} \sqrt{n}$ is convergent or divergent.**

We note that the series $\sum_{n = 1}^{\infty} \sqrt{n} = \sum_{n=1}^{\infty} n^{1/2} = \sum_{n=1}^{\infty} \frac{1}{n^{-1/2}}$. Thus $p = -1/2 ≤ 1$ and by the p-Series test the series $\sum_{n = 1}^{\infty} \sqrt{n}$ is divergent.