TITLE: For what values for m does $\sum \limits_{k=2}^{\infty}\frac{1}{(\ln{k})^m}$ converges?
QUESTION [7 upvotes]: For what values for m does $$\sum \limits_{k=2}^{\infty}\frac{1}{(\ln{k})^m}$$ converge?
What about 
$$\sum_{k=2}^{\infty}\frac{1}{(\ln(\ln{k}))^m}$$ or more generally
$$\sum_{k=2}^{\infty}\frac{1}{(\ln(\cdots (\ln{k}))\cdots)^m}$$ ?

REPLY [4 votes]: For the first series,we can use Cauchy's condensation criterion when $p>0$ (when $p\leq 0$ the series is divergent): 

If $\{a_n\}$ is a decreasing sequence of positive numbers, the series $\sum_ka_k$ is convergent if and only if $\sum_k2^ka_{2^k}$ is convergent. 

We have with $a_k=\frac 1{(\ln k)^m}$ that $2^ka_k=\frac{2^k}{k^m(\ln 2)^m}$ and it diverges. 
An alternative way is to note that $\ln k\leq k^{1/m}$ for a $m$ large enough (which depend on $m$), and use the divergence of harmonic series. 
We have for $x\geq 1$ that $\log x\leq x$. Denote $f_N$ the $N$-th iterate of the logarithm. For a fixed $N$, we can find $n(N)$ such that for $k\geq n(N)$, $f_{N-1}(k)\geq 1$. Hence $f_N(k)\leq \log k$ for $k$ large enough, which shows that the series is divergent for all $m$ and all $N$.
The problem here is that the logarithm grows too slowly. Taking iterates doesn't make the thing better, and the exponent doesn't change anything.