# Convergence\Divergence of $\sum_{k=2}^{\infty} \frac{\cos(\ln(\ln k))}{\ln k}$

Test for convergence the series $$\sum_{k=2}^{\infty} \frac{\cos(\ln(\ln k))}{\ln k}$$ My first thought was related to the use of the integral test, but things seem hard.
Could we resort here to some nicers tools? Thanks

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## 1 Answer

Pick an integer $n$ and consider all the terms with $\ln \ln k \in [2n\pi-1 ; 2n\pi+1]$. Then $\cos(\ln(\ln k)) \ge \cos 1$ and $\ln k \le e^{2n\pi+1}$, so $a_k \ge (\cos 1) e^{-2n\pi-1}$.
Next, we want to estimate how many $k$ have $\ln \ln k \in [2n\pi-1 ; 2n\pi+1]$. This is equivalent to $k \in [e^{e^{2n\pi-1}};e^{e^{2n\pi+1}}]$, whose length is greater than $C e^{e^{2n\pi+1}}$ for some $C>0$ (because $e^{e^{2n\pi-1}}$ is negligible in front of this).
Therefore the sum of those consecutive terms is greater than $C e^{e^{2n\pi+1}}(\cos 1)e^{-2n\pi-1}$. And since $e^x/x$ gets arbitrarily big as $x$ gets larger, you deduce that the sum of those consecutive terms can be arbitrarily large if you choose $n$ high enough.

As a result, the sequence $\sum_{k=0}^n a_k$ is not a Cauchy sequence, and doesn't converge.

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+1. And this nearly shows that the closure of the set of partial sums is the whole real line. – Did Dec 17 '12 at 16:50
@mercio: Very nice. Thanks! – I'm an artist Dec 17 '12 at 17:00
Really good. I tried that approach myself but couldn't make it work. – Martin Argerami Dec 17 '12 at 21:00