For what value of $p$ would the series $\sum_{n=2}^\infty \frac{1}{n^p\ln(n)}$ converge?

Question

For what value of $p$ would the series $\sum_{n=2}^\infty \frac{1}{n^p\ln(n)}$ converge?

I have tried using the series convergence test, and limit comparison test, which does't help. So i am considering using the divergence test, and finding all values for p for which the limit is not $0$

Answer 1

Interestingly, $\sum_{n=2}^\infty \frac{1}{n^pln(n)} $ converges exactly when $\sum_{n=2}^\infty \frac{1}{n^p} $ converges: converges for $p > 1$ and diverges for $p \le 1$.

The comparison test does it for $p > 1$ and the fact that $\dfrac{\ln(n)}{n^c} \to 0$ for any $c > 0$ does it for $p < 1$.

The only problem is when $p=1$ and this can be handled by either the integral test or Cauchy's condensation test.