Use the integral test to determine whether the series

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

converges of diverges.


"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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