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

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

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.