How to prove the series $\sum_{\ n=2}^{\infty} \frac{1}{(\ln n)^k}$ diverge for all k greater than zero? My idea is to show that there is some N such that for all n greater than N, $\frac{1}{(\ln n)^k} >   \frac{1}{n}$ , hence from the Nth term onward the series is greater than harmonic series. Is there any better proof?
 A: As the series' sequence is monotone descending to zero, you can use Cauchys' Condensation Test, and then:
$$\frac{2^n}{(\log(2^n))^k}=\frac{2^n}{n^k\log^kn}$$
and now use, say the quotient test:
$$\frac{2^{n+1}}{(n+1)^k\log^k(n+1)}\frac{n^k\log^kn}{2^n}=2\left(\frac n{n+1}\right)^k\left(\frac{\log n}{\log(n+1)}\right)^k\xrightarrow[n\to\infty]{}2\cdot1\cdot1=2>1$$
and thus the series diverges.
I can't say whether this is "better", but it uses basic things with basic computations...
A: In my opinion your approch is fine. It suffices to consider the limit $$\lim_{n\to +\infty}\frac{(\ln n)^k}{n}=0$$
Then, by definition of limit, there exists an integer $N$ such that for $n>N$, $\frac{(\ln n)^k}{n}<1$, that is $\frac{1}{n}<\frac{1}{(\ln n)^k}$.
P.S. The above limit is zero because for $k>0$,
$$\lim_{x\to +\infty}\frac{(\ln x)}{x^{1/k}}=\lim_{x\to +\infty}\frac{x^{-1}}{\frac{x^{1/k-1}}{k}}=\lim_{x\to +\infty}kx^{-1/k}=0$$
where we used Hopital.
A: I suppose that you could use the standard tests.
If $$u_n=\frac 1 {(\log(n))^k}$$ $$\frac{u_{n+1}}{u_n}=\left(\frac{\log(n)}{\log(n+1)}\right)^k$$ Using Taylor series for large values of $n$, then$$\frac{u_{n+1}}{u_n}=1-\frac{k}{n \log \left({n}\right)}+O\left(\frac{1}{n^2}\right)$$  which is inconclusive.
Now, using Raabe's test and doing the same 
$$n\left(\frac{u_n}{u_{n+1}}-1\right)=\frac{k}{ \log \left({n}\right)}+O\left(\frac{1}{n}\right)$$ which is $\lt 1$. So, divergence. 
