I have problem with these two:

a) $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{\ln{n}})^{\ln{n}}}$

b) $\displaystyle \sum_{n=3}^{\infty} \frac{1}{n \cdot \ln{n} \cdot \ln{\ln{n}}}$

My try: a) I come to $S_n=(n^{\ln{\ln{\ln{n}}}})^{-1}$ but I don't know what next.

b) using condenstaion test I get $S_n=(n \cdot \ln{2} \cdot \ln{(n\cdot\ln2)})^{-1}$



a) rewrite the denominator as $n^{\ln(\ln(\ln n))}$, and then compare to $\sum \frac{1}{n^2}$

b) try using the Integral Test:

Evaluate $\int_3^{\infty} \frac{1}{x\ln x\ln(\ln x)} dx $ by letting $u=\ln(\ln x)$ and $du=\frac{1}{x\ln x} dx$

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    $\begingroup$ For (b) shouldn't it be with $\int_0^x \frac{du}{u \ln u \ln \ln u} = \ln\ln\ln x$? $\endgroup$ – Clement C. Dec 21 '14 at 18:03
  • 1
    $\begingroup$ @ClementC. I will try to edit my answer to make it clearer what I meant. $\endgroup$ – user84413 Dec 21 '14 at 18:48

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