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.


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.  

