Why does the series converge I know that the series $\sum_{n=2}^{\infty} \dfrac{{log(n^b)}^{M}}{n^b}$, where 1 < b < $\infty$ and M is an arbitrary nonnegative integer converges, but I don't know what I should compare the series to. Thanks for any help in advance.

I personally think the more efficient way to show the convergence of the series is via Cauchy condensation test which states: 

(Cauchy's Condensation Test)   For a non-negative non-increasing sequence $\{a_n\}$, the series
  $\sum a_n$ converges iff $\sum 2^k a_{2^k}$ converges.

Observe
\begin{align}
\sum^\infty_{k=1} 2^k \frac{(\log 2^{bk})^M}{2^{bk}} = \sum^\infty_{k=1}\frac{(bk \log 2)^M}{2^{(b-1)k}}=(b\log 2)^M\sum^\infty_{k=1}\frac{k^M }{2^{(b-1)k}} <\infty
\end{align}
which converges iff  $b>1$.
