How to determine convergence of the series $\sum\limits_{n=2}^{\infty}\frac{\cos(\ln(\ln(n))}{\ln(n)}$ ?

I tried to get somewhere with Integral criteria and with comparing to other series but didn't get much done.

Do you know with which series should I compare it to?

  • $\begingroup$ You have to start from $n=2$, for $n=1$ denominator is 0 and numerator is undefined. $\endgroup$
    – Wojowu
    Jun 25 '15 at 20:37
  • 2
    $\begingroup$ Hint: Fix some large $N$ and evaluate the sum $S_N$ of the terms such that $$2N\pi-\pi/3\leqslant\ln\ln n\leqslant2N\pi+\pi/3.$$ For every such $n$, the cosine is at least $1/2$ and $\ln n\leqslant ce^{2N\pi}$ hence $S_N\geqslant c'e^{-2N\pi}T_N$ where $T_N$ is the number of terms in $S_N$. Can you estimate $T_N$ and conclude? $\endgroup$
    – Did
    Jun 25 '15 at 20:44
  • $\begingroup$ Shouldn't then be $T_N = N$ because we are summing by the fixed integer $N$? From this I would say that the series diverges. $\endgroup$
    – purgerica
    Jun 25 '15 at 20:58
  • 1
    $\begingroup$ Huh? $T_N=\#\{n\mid2N\pi-\pi/3\leqslant\log\log n\leqslant2N\pi+\pi/3\}$ makes $T_N$ quite different from $N$. $\endgroup$
    – Did
    Jun 25 '15 at 21:24

Just expanding Did's comment, let we take a huge $N\in\mathbb{N}$ and let $A_N\subset\mathbb{N}$ be the set of integers such that: $$ 2\pi N -\frac{\pi}{3}\leq \log\log n \leq 2\pi N+\frac{\pi}{3} \tag{1}$$ holds. For every $n\in A_N$ we have $\cos\log\log n\geq\frac{1}{2}$ and $\log n\leq e^{\pi/3}\cdot e^{2\pi N}$. However,

$$ |A_N|\geq \exp\exp\left(2\pi N+\frac{\pi}{3}\right)-\exp\exp\left(2\pi N-\frac{\pi}{3}\right)-2 \tag{2}$$ so $$ \sum_{n\in A_N}\frac{\cos\log\log n}{\log n} \tag{3} $$ grows unbounded, and the sequence given by: $$ S_M = \sum_{n=2}^{M}\frac{\cos\log\log n}{\log n} \tag{4}$$ is not a Cauchy sequence, hence the original series is not convergent.


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