Study the convergence of $\sum ^{\infty }_{n=2}\frac{1}{n\ln n\cdot \left( \ln \ln n\right) ^{2}}$ by using the Cauchy Condensation Test This is what I did so far.
We need to check the convergence (divergence) of $$\sum ^{\infty }_{n=2}2^{n}\frac{1}{2^{n}\ln 2^{n}\left( \ln \ln 2^{n}\right) ^{2}}=\sum ^{\infty }_{n=2}\frac{1}{n\ln 2\cdot \ln ^{2}\left( n\ln 2\right) }$$
But I don't understand how to continue from this point. I tried applying the Cauchy condensation test for the second time but didn't get anything helpful.
Any hint on how to continue?
Continued.
After applying test for the second time I got
$$\sum ^{\infty }_{n=2}2^{n}\frac{1}{2^{n}\ln 2\left( \ln 2^{n}+\ln \ln 2\right) ^{2}}=\sum ^{\infty }_{n=2}\frac{1}{\ln 2\left( n\ln 2+\ln \ln 2\right) ^{2}}$$
From here we can say that.
$$\sum ^{\infty }_{n=2}\frac{1}{\ln 2\left( n\ln 2+\ln \ln 2\right) ^{2}}\leq\sum ^{\infty }_{n=2}\frac{1}{\ln 2\left( n\ln 2\right) ^{2}}$$
We can tell that the series $\dfrac{1}{(\ln 2)^3}\sum\limits ^{\infty }_{n=2}\dfrac{1}{n^{2}}$ do converge, so we can draw conclusions from here.
Is this correct?
 A: Apply again Cauchy  condensation test to $\sum_{n=2}^\infty \frac{1}{n \log^2 n}$:
$$\sum_{n=2}^\infty 2^n\frac{1}{2^n \log^2 2^n} = \sum_{n=2}^\infty \frac{1}{ n^2\log^2 2}$$ converges. Hence by Cauchy  condensation test $\sum_{n=2}^\infty \frac{1}{n \log^2 n}$ is convergent and the initial series too.
A: Note that $n\ln 2\ge \sqrt n$ for all $n\ge 3$, so the given series becomes
\begin{align}
\sum ^{\infty }_{n=2}\dfrac{1}{n\ln 2\cdot \ln ^{2}\left( n\ln 2\right) }&\le\dfrac{1}{2\ln2\ln^2(2\ln2)}+\dfrac{1}{\ln2}\sum_{n=3}^\infty\dfrac1{n\ln^2(\sqrt n)}\\
&=\dfrac{1}{2\ln2\ln^2(2\ln2)}+\dfrac{4}{\ln2}\sum_{n=3}^\infty\dfrac1{n\ln^2(n)}
\end{align}
Now applying Cauchy Condensation test or integral test gives you convergence directly (Note that integral test would have given you convergence for the original sequence directly anyway).
A: When $n\ge 2$ and $\ln n>|\ln \ln 2|/2$ we have $(\ln (n\ln 2))^2=(\ln n +\ln\ln 2)^2>((\ln n)/2)^2=(\ln n)^2/4.$
So all but finitely many terms in the series on the RHS of your first line are less than the terms of $\sum_{n\ge 2} [(n\ln 2)(\ln n)^2/4]^{-1}.$ Apply the Condensation Test to this series.
Actually we can say "all terms" instead of "all but finitely many terms" because $n\ge 2\implies \ln n>|\ln \ln 2|/2$ but we don't need to know that.
