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The series is: $\displaystyle\sum_{n=1}^\infty \dfrac1{n (\ln n)^p}$

I don't know what to do from here since $p$ is on $\ln$.

Would $p$ still have to be $> 1$ since $\ln$ is changing in terms of $n$?

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This is really classic : Bertrand series. –  Krokop Aug 17 at 23:12

3 Answers 3

up vote 1 down vote accepted

$\textbf{Hint:}$ Use Integral test for convergence.

By Integral test of convergence you calculate the integral $\displaystyle \int_{1}^{\infty} \frac{1}{x(\ln x)^p}dx=\int_{0}^{\infty} \frac{1}{t^p} dt$ (substitution $t=\ln x$).

For $p=-1$ you have $\displaystyle \int \frac{1}{t^p} dt=\log|t|+C$, so integral diverges.

For $p \neq -1 $ you have $\displaystyle \int \frac{1}{t^p} dt=\frac{p-1}{t^{p-1}}+C$, so for $p<-1$ integral diverges, for $p>-1$ converges.

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This is clearly not true. If $p = -\frac{1}{2}$ then $\frac{1}{n(\ln{n})^p} = \frac{\ln{n}^{\frac{1}{2}}}{n} > \frac{1}{n}$ (for $n > 3$) so the series must diverge. –  Davis Yoshida Aug 17 at 22:53
    
That should say "for $p < 1$ the integral diverges, for $p > 1$ converges". –  JimmyK4542 Aug 17 at 23:00

Hint: By the integral test, $\displaystyle\sum_{n=1}^\infty \dfrac1{n (\ln n)^p}$ converges iff $\displaystyle\int_{1}^{\infty}\dfrac{dx}{x(\ln x)^p}$ converges.

Now substitute $u = \ln x$ into that integral, and see what you get.

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The Cauchy condensation test easily gives that your series converges iff $p>1$.

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