For what values of p and q will $\sum\limits_{n=2}^{\infty }{ \frac{1}{n^p\times(\ln(n))^q} }$ converge?

For sufficiently large $n$ independent of $p$ and $q$, $n > ln(n)$, so $n^p$ is dominant.

1. $p\gt 1$,
   $\sum\limits_{n=2}^{\infty }{ \frac{1}{n^p\times(\ln(n))^q} } \le \sum\limits_{n=2}^{\infty }{ \frac{1}{n^p}}$
   $\sum\limits_{n=2}^{\infty }{ \frac{1}{n^p}}$ converges, so $\sum\limits_{n=2}^{\infty }{ \frac{1}{n^p\times(\ln(n))^q} }$ converges for all $q$.

2. $p=1$,
   $\sum\limits_{n=2}^{\infty }{ \frac{1}{n}}$ diverges(it’s harmonic numbers).
   But using integral test, $\int_{2}^{\infty}{\frac{1}{x\times(\ln{x})^q}dx}=\left[\frac{1}{(-q+1)\times(\ln{x})^{q-1}}\right]^\infty_2$, range of $q$ can be divided.

   • $q \gt 1$
     $\sum\limits_{n=2}^{\infty }{ \frac{1}{n^p\times(\ln(n))^q} }=\sum\limits_{n=2}^{\infty }{ \frac{1}{n\times(\ln(n))^q} }$ converges.
   • $q \lt 1$
     $\sum\limits_{n=2}^{\infty }{ \frac{1}{n^p\times(\ln(n))^q} }=\sum\limits_{n=2}^{\infty }{ \frac{1}{n\times(\ln(n))^q} }$ diverges.
   • $q = 1$
     $\int_{2}^{\infty}{\frac{1}{x\times(\ln{x})^q}dx}=\left[\ln{x}\right]^\infty_2$ diverges.

3. $p\lt1$,
   $\sum\limits_{n=2}^{\infty }{ \frac{1}{n^p\times(\ln(n))^q} } \ge \sum\limits_{n=2}^{\infty }{ \frac{1}{n^p}}$
   $\sum\limits_{n=2}^{\infty }{ \frac{1}{n^p}}$ diverges, so $\sum\limits_{n=2}^{\infty }{ \frac{1}{n^p\times(\ln(n))^q} }$ diverges for all $q$.