Why does the series converge

Question

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.

Answer 1

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.

Answer 2

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 $b>1$.