# Science:Infinite Series Module/Units/Unit 2/2.2 The Integral Test/2.2.07 The p-series

## Problem

For what values of p does the infinite series

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{p}}}}$

converge?

## Complete Solution

### Step (1): Consider p > 0 and p ≠ 1

When p > 0 and p ≠ 1, the function

${\displaystyle f(x)={\frac {1}{x^{p}}}}$

is continuous, decreasing, and positive when x is in the interval [1,∞). Using the integral test,

{\displaystyle {\begin{aligned}\int _{1}^{\infty }{\frac {1}{x^{p}}}dx&=\lim _{b\rightarrow \infty }\int _{1}^{b}{\frac {1}{x^{p}}}dx\\&=\lim _{b\rightarrow \infty }{\frac {1}{-(p-1)x^{p-1}}}{\Big |}_{1}^{b}\\&={\begin{cases}\infty &{\text{if }}p<1,\\{\frac {1}{p-1}}&{\text{if }}p>1\end{cases}}\end{aligned}}}

Therefore, the infinite series converges when p > 1, and diverges when p is in the interval (0,1).

### Step (2): Consider p ≤ 0 and p = 1

If p=1, then we have the harmonic series

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}}$

which we know diverges.

If p ≤ 0, the infinite series diverges (by the divergence test).

Therefore, the given series only converges for p > 1.

## The p-Series

The result of this example can be summarized as follows.

The p-Series

The p-series

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{p}}}}$

is convergent if p > 1 and divergent if p ≤ 1.

Much like a geometric series, we can use this result to determine whether a given infinite series converges by inspection. For example, the infinite series

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{\sqrt {1+k}}}}$

diverges because it is a p-series with p equal to 1/2 (you may want to let u=(1+k) to see this).