The Divergence and Integral Tests: Learn It 3

The p-Series

The harmonic series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex] and the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex] are both examples of a type of series called a [latex]p[/latex]-series.

[latex]p[/latex]-series

For any real number [latex]p[/latex], the series

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{p}}[/latex]

 

is called a [latex]p[/latex]-series.

We know the [latex]p[/latex]-series converges when [latex]p = 2[/latex] and diverges when [latex]p = 1[/latex]. What happens for other values of [latex]p[/latex]? While computing exact values of most [latex]p[/latex]-series is difficult or impossible, we can determine their convergence behavior.

Testing Different Values of [latex]p[/latex]

Case 1: [latex]p \leq 0[/latex]

  • If [latex]p < 0[/latex], then [latex]\frac{1}{n^p} \to \infty[/latex] as [latex]n \to \infty[/latex].
  • If [latex]p = 0[/latex], then [latex]\frac{1}{n^p} = 1 \to 1[/latex] as [latex]n \to \infty[/latex].

By the divergence test, [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}[/latex] diverges if [latex]p \leq 0[/latex].

Case 2: [latex]p > 0[/latex]

When [latex]p > 0[/latex], the function [latex]f(x) = \frac{1}{x^p}[/latex] is positive, continuous, and decreasing. We can apply the integral test by comparing:

[latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}[/latex] and [latex]\int_1^{\infty} \frac{1}{x^p}dx[/latex].

For [latex]p > 0, p \neq 1[/latex]:

[latex]{\displaystyle\int }_{1}^{\infty }\frac{1}{{x}^{p}}dx=\underset{b\to \infty }{\text{lim}}{\displaystyle\int }_{1}^{b}\frac{1}{{x}^{p}}dx=\underset{b\to \infty }{\text{lim}}\frac{1}{1-p}{x}^{1-p}{|}_{1}^{b}=\underset{b\to \infty }{\text{lim}}\frac{1}{1-p}\left[{b}^{1-p}-1\right][/latex].

The key insight is how [latex]b^{1-p}[/latex] behaves:

  • [latex]b^{1-p} \to 0[/latex] if [latex]p > 1[/latex]
  • [latex]b^{1-p} \to \infty[/latex] if [latex]p < 1[/latex]

Therefore:

[latex]{\displaystyle\int _{1}^{\infty}} \dfrac{1}{x^{p}}dx = \Bigg\{ \begin{array}{c} \frac{1}{p-1}\text{ if }p>1\\ \infty \text{ if }p<1\end{array}[/latex]

This means [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}[/latex] converges if [latex]p > 1[/latex] and diverges if [latex]0 < p < 1[/latex].

[latex]p[/latex]-series convergence test

The [latex]p[/latex]-series [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}[/latex] converges if and only if the exponent [latex]p > 1[/latex].

[latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p} \begin{cases} \text{converges} & \text{if } p > 1 \\ \text{diverges} & \text{if } p \leq 1 \end{cases}[/latex]

For each of the following series, determine whether it converges or diverges.

  1. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{4}}[/latex]
  2. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{\frac{2}{3}}}[/latex]