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## Problem

Use the integral test to determine whether the series

$\sum_{n=1}^{\infty} \frac{\ln^3(n)}{n^6}$

converges of diverges.

## Solution

"Convert" the series in an integral. If the integral converges, then the series converges. If the integral diverges, then the series diverges.

Let $y = \ln(x)$, then $x = e^y$ and $dy = \frac{1}{x} dx$.

By the method of substitution, the integral becomes

$\int_{0}^{\infty} y^3 {e}^{-5y} dy$.

One can use integration by parts several times or use the Gamma function to evaluate this integral. I will use the Gamma function.

$\int_{0}^{\infty} t^n {e}^{-at}dt = \frac{\Gamma (n+1)}{{a}^{n+1}} = \frac{n!}{{a}^{n+1}}$

This means

$\int_{0}^{\infty} y^3 {e}^{-5y} dy = \frac{3!}{5^4}$
$\int_{0}^{\infty} y^3 {e}^{-5y} dy = \frac{6}{625}$

Since the integral converges, the above infinite series is convergent.

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