I need to show that the series $$\sum_{n=3}^{\infty} \frac{(\log(\log(n))^2}{n(\log(n))^2}$$ converges.

I'm trying to find some upper bound for it, since the tests i've used so far did not lead to any useful conclusion. I think that a good upper bound might be the series $\sum_{n=3}^{\infty} \frac{1}{n(\log(n))^{1+\epsilon}}$, for $\epsilon >0$ because then i could use the fact that the series $\sum_{n=3}^{\infty} \frac{1}{n^{1+\epsilon}}$ converges. Therefore i tried to approximate the expression $(\log \log n)^2$ to something like

$$(\log \log n)^2 \le (\log n)^{1+\delta}$$

for some $\delta > 0$ but without success. Could you please help me?


Hint: for $n$ big enough, $$ (\log \log n)^2 < \sqrt{\log n } $$ Now use the fact that $$ \sum \frac 1{n(\log n)^{3/2}}<\infty $$

  • $\begingroup$ Thank you for your time. Are you sure that $(\log \log n)^2 < \sqrt{\log n}$? Am i misinterpreting something? I tried to plot both functions, but i get $(\log \log n) > \sqrt{\log n}$ for $n$ large enough. $\endgroup$ – Bman72 Nov 16 '14 at 11:12
  • $\begingroup$ Yes i am sure. Put n=e^e^t before plot. $\endgroup$ – mookid Nov 16 '14 at 11:24

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