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.
 A: 
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequality

$$\bbox[5px,border:2px solid #C0A000]{\log(x)\le x-1<x} \tag 1$$ 
for $x>0$.

Using $(1)$ along with the property that $\log(x^a)=a\log(x)$, we find that for $a\ge 0$
$$\log(x)\le \frac{x^a}{a}$$
For any $M\ge0$, choose $a$ such that $aM-b<-1$ (or, $a<\frac{b-1}{M}$).  Then, we have
$$ \sum_{n=2}^N\frac{\log^M(n^b)}{n^b}\le \left(\frac ba\right)^M \sum_{n=2}^Nn^{aM-b} \tag 2$$

By the comparison test (or integral test), the sum on the right-hand side of $(2)$ converges and hence by comparison the integral on the left-hand side converges also.  

A: 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  $b>1$.
