How to decide about the convergence of $\sum(n\log n\log\log n)^{-1}$?

Solution 1:

According to Rudin's criterion (or whatever its name is) $$\sum_{n=3}^\infty \frac{1}{n \log n (\log \log n)^p}$$ converges iff $$ \sum_{k=2}^\infty \frac{2^k}{2^k \log (2^k) (\log \log (2^k))^p} $$ converges or equivalently if $$ \sum_{k=2}^\infty \frac{1}{k\log 2 (\log (k\log 2))^p}= \frac{1}{\log 2}\sum_{k=2}^\infty \frac{1}{k (\log k+\log\log 2)^p} $$ converges. But $$ \log k\ge \log k+\log\log 2\ge \frac{1}{3}\log k, $$ and hence the sum $$ \frac{1}{\log 2}\sum_{k=2}^\infty \frac{1}{k (\log k+\log\log 2)^p}$$ converges iff $$ \frac{1}{\log 2}\sum_{k=2}^\infty \frac{1}{k (\log k)^p} $$ converges, which we already know that it converges iff $p>1$.

Solution 2:

Using this method which's called Cauchy condensation we get

$$\sum_{k\ge1}\frac{2^k}{2^k\ln 2^k\ln\ln(2^k)}=\frac1{\ln2}\sum_{k\ge1}\frac1{k\ln(k\ln2)}\sim\frac1{\ln2}\sum_{k\ge1}\frac1{k\ln(k)}$$ so the series

$$\sum_{n\ge1}\frac1{n\ln n\ln(n\ln n)}$$ is divergent. Can you now solve the second series?