Convergence of $\sum_{n=2}^\infty \frac{1}{n\ln n(\ln(\ln n))^2}$

Prove the series $$\sum_{n=2}^\infty \frac{1}{n\ln n(\ln(\ln n))^2}$$ converges.

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Since the general term is a positive and decreasing sequence, by the integral test, the series behaves as the following integral: $$\int_3^\infty \frac{dx}{x\ln x (\ln(\ln x))^2}.$$By a change of variable $$u=\ln(\ln x),\quad du=\frac{dx}{x\ln x},$$ we have $$\int_3^\infty \frac{dx}{x\ln x (\ln(\ln x))^2}=\int_{\ln(\ln 3)}^\infty \frac{du}{u^2}=-\frac{1}{u}\bigg|_{\ln(\ln 3)}^{\infty}=\frac{1}{\ln(\ln 3)}.$$ So it is convergent. [We start with $n=3$ to avoid the pole at $x=e$ where the integral may diverge.]

We can also apply the Cauchy condensation test. $$\sum_n a_n\;\text{converges iff}\;\sum_n 2^n a_{2^n}\;\text{converges}.$$ Setting $a_n=\dfrac{1}{n\ln n(\ln(\ln n))^2}$, we have $$\sum_n 2^n a_{2^n}=\sum_{n=2}^\infty \frac{2^n}{2^n(n\ln 2)(\ln n+\ln(\ln 2))^2}\leq \frac{1}{\ln 2}\sum_{n=2}^\infty \frac{1}{n\ln^2 n}.$$ Similarly, for $\sum\limits_{n=2}^\infty\dfrac{1}{n\ln^2 n}$, we have $$\sum_{n=2}^\infty \frac{2^n}{2^n\ln^2(2^n)}=\frac{1}{\ln^2 2}\sum_{n=2}^\infty \frac{1}{n^2}.$$ So the series is convergent.