# How do you use the Integral test on the infinite series sum_(n=1)^oo1/n^5 ?

Sep 24, 2014

By Integral Test,

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 5$ converges.

Let us look at some details.

Let us evaluate the corresponding improper integral.

${\int}_{1}^{\infty} \frac{1}{x} ^ 5 \mathrm{dx}$

$= {\lim}_{t \to \infty} {\int}_{1}^{t} {x}^{- 5} \mathrm{dx}$

$= {\lim}_{t \to \infty} {\left[{x}^{- 4} / - 4\right]}_{1}^{t}$

$= - \frac{1}{4} {\lim}_{t \to \infty} {\left[\frac{1}{x} ^ 4\right]}_{1}^{t}$

$= - \frac{1}{4} {\lim}_{t \to \infty} \left[\frac{1}{t} ^ 4 - 1\right]$

$= - \frac{1}{4} \left(0 - 1\right) = \frac{1}{4}$

Since the integral

${\int}_{1}^{\infty} \frac{1}{x} ^ 5 \mathrm{dx}$

converges to $\frac{1}{4}$,

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 5$

also converges by Integral Test.