Determine the convergence or divergence of $\sum_{2}^{\infty}\frac{1}{(\log n)^{s}}$, where $s \in \mathbb{R}$ is given.

Question

Since $$\frac{1}{(\log n)^{s}} > \frac{1}{n^{s}}$$ for large $n$, if $s \leq 1$ then $\sum_{2}^{\infty}\frac{1}{(\log n)^{s}}$ diverges.

But for $s > 1$ I have not yet figured out a proof.

Answer 1

A different proof. $$ \lim_{n\to\infty}\frac{(\log n)^s}{n}=0\quad\forall s>0. $$ Thus $$ \frac{1}{(\log n)^s}\ge\frac{1}{n}\quad\text{for $n$ large enough (depending on $s$.)} $$

Answer 2

There is a theorem that states that if $a_n$ is a decreasing sequence, then $\sum_n a_n$ converges iff $\sum_i 2^i a_{2^i}$ converges.

See chapter 3 of Baby Rudin for the proof.

Applying that here we get $$\sum_i \frac{2^i}{(\log(2^i))^s} = \frac{1}{(\log 2)^s} \sum_i \frac{2^i}{i^s}$$

which diverges, hence the original series diverges.