p-Series

Calculus 2

p-Series Test

A p-series is a series of the form:

\sum_{n=1}^\infty \frac{1}{n^p} = \frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + \cdots

where $p$ is a positive constant.

The $p$ -series formed when $p=1$ is known as the harmonic series and diverges:

\sum_{n=1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots

The p-Series Test states:

If $p > 1$ , the $p$ -series $\sum_{n=1}^\infty \frac{1}{n^p}$ converges.
If $0 < p \leq 1$ , the $p$ -series $\sum_{n=1}^\infty \frac{1}{n^p}$ diverges.

Note: We can use the similarity (or difference) in conditions for the p-Series Test to convince ourselves of the conditions for the Geometric Series Test and vice versa.

Example 1

Determine the convergence/divergence of the series $\sum_{n=1}^\infty \frac{1}{n^3}$ using the $p$ -Series Test.

View Answer

Converges

View Solution

Since $p = 3 > 1$ , the series converges by the p-Series Test.

Example 2

Determine the convergence/divergence of the series $\sum_{k=1}^\infty \frac{1}{\sqrt{k}}$ using the $p$ -Series Test.

View Answer

Diverges

View Solution

Since $p = \frac{1}{2} \leq 1$ , the series diverges by the p-Series Test.

Example 3

Determine the convergence/divergence of the series $\sum_{n=1}^\infty \frac{5}{n^{-2}}$ using the $p$ -Series Test.

View Answer

Inconclusive (not a $p$ -series)

View Solution

Recall that $p$ must be a positive constant. Therefore, this series is not considered a $p$ -series and the $p$ -Series Test is inconclusive.

However, we could still determine the convergence/divergence of this series using tests we've already discussed.

Rewriting the series, we get $\sum_{n=1}^\infty \frac{5}{n^{-2}} = \sum_{n=1}^\infty 5n^2$ . Using the nth Term Test, we see that this series diverges.

Example 4

Determine the convergence/divergence of the series $\sum_{n=1}^\infty \frac{12\sqrt{n}}{n^{3/2}}$ using the $p$ -Series Test.

View Answer

Diverges

View Solution

Simplifying the series, we get $\frac{12\sqrt{n}}{n^{3/2}} = \frac{12n^{1/2}}{n^{3/2}} = \frac{12}{n}$ .

Since $p = 1$ , the series diverges by the p-Series Test.

Geometric Series

Comparison Tests