# Decide convergence of the series .

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 $\sum \frac{1}{n^2}$
Evaluate $\int_3^{\infty} \frac{1}{x\ln x\ln(\ln x)} dx$ by letting $u=\ln(\ln x)$ and $du=\frac{1}{x\ln x} dx$
• For (b) shouldn't it be with $\int_0^x \frac{du}{u \ln u \ln \ln u} = \ln\ln\ln x$? – Clement C. Dec 21 '14 at 18:03