Prove sum of $ 1 / (\ln(n)^s) $diverges for $s>1$ Hi i need help with proving that the sum of:
$\sum_{n=2}^{\infty}1/(\ln(n)^s)$ is divergent for all $s>1$.
I have tried to prove various limits that might help me prove this is divergent but with no success. 
Any tips?
 A: You can use Cauchy's Condensation Test ,and then
$$\frac{2^n}{(\log 2^n)^s}=\frac{2^n}{n^s\log^s2}$$
and for this last apply the $\;n\,-$ th root test:
$$\sqrt[n]\frac{2^n}{n^s}=\frac2{(\sqrt[n] n)^s}\xrightarrow[n\to\infty]{}2>1$$
so our series diverges...and observe that we do not care what is $\;s\;$ ...
A: Hint: $(\ln n)^s < n$ for large enough $n$ so $\frac{1}{(\ln n)^s} > \frac{1}{n}$
A: The trick is to do two things:


*

*Maintain benchmark sums that you know how they work (e.g. $\sum n^{-p}$)

*When encountering a new sum, ask yourself: "How does this new function compare to the old ones?"


In this case, you (should) know the harmonic series as a benchmark, so let's see if we can compare $\frac{1}{\ln(n)^s}$ to $\frac{1}{n}$. That's the same as comparing $\ln(n)^s$ to $n$.
But we know that $\ln(n)$ will always lose to any power function, so for large enough $n$, we have $\ln(n)^s < n$. Thus the tail of the sum $\sum \ln(n)^{-s}$ is bounded below by the tail of the sum $\sum n^{-1}$ and it must diverge.
To make this intuition precise, use the limit comparison theorem.
A: Ok let's go for a too complicated proof which manages to exhibit an actual value for $n$.

Let $s \geq 1,\ s = a+r$ where $a = \lfloor s \rfloor$ and $r = \{s\}$.
Claim :
  $$n > \exp\left( ((a+2)!)^{1/(2-r)}\right) \implies \ln(n)^s < n
$$
  $\tiny{\text{(which is certainly one of the worse bound ever)}}$

Proof : Let us define $n = e^m$.
$$
\begin{align}
\ln(n)^s &< n &\Longleftrightarrow \\
m^s &< e^m = \sum_{i=0}^\infty \frac{m^i}{i!} &\overset{(*)}{\Longleftarrow} \\
m^{a+r} &< \frac{m^{a+2}}{(a+2)!} &\Longleftrightarrow \\
(a+2)! &< m^{r-2} &\Longleftrightarrow \\
((a+2)!)^{1/(r-2)} &< m &\Longleftrightarrow \\
\exp\left(((a+2)!)^{1/(r-2)}\right) &< e^m = n
\end{align} 
$$
$(*)$ Where we only retain the term $i = a+2$
This shows that at some point $\frac{1}{\ln(n)^s} > \frac1n$ and by comparison with the harmonic series, your series diverges.
