Prove the series $\sum_{n=2}^\infty \frac{1}{n \ln(n)(\ln(\ln(n)))^2}$ converges I need to prove the series $$\sum\limits_{n=2}^\infty \frac{1}{n \ln(n)(\ln(\ln(n)))^2}$$ converges.
I cant use comparison test, since if I take away any term in denominator I know it will diverge. Any help please?
 A: Hint. Since the general term is a positive and decreasing sequence, by the integral test, your series behaves as the following integral:
$$
\int_3^{\infty}\frac{{\rm d} x}{x \ln x \:(\ln(\ln x))^2}. \tag1
$$ By the change of variable $$u=\ln(\ln x), \quad du=\dfrac{dx}{x \ln x },$$ we easily have

$$
\begin{align}
\int_3^{\infty}\frac{{\rm d} x}{x \ln x \:(\ln(\ln x))^2} \mathrm dx=\int_{\ln(\ln 3)}^{\infty}\frac{{\rm d} u}{u^2}
=\left[ -\frac{1}{u}\right]_{\ln(\ln 3)}^{\infty}
=\frac1{\ln(\ln 3)}.
\end{align}
$$

Your initial series is convergent.
Edit. Thanks to user109899's remark, we have singled out the term coming from $n=2$.
A: Hint. You may apply the Cauchy condensation test since the sequence is positive and decreasing:

$\sum_n a_n$ converges if and only if $\sum_n 2^n \:a_{2^n}$ converges. 

By setting $\displaystyle a_n=\frac{1}{n \ln(n)(\ln(\ln(n)))^2} $, we easily get
$$\sum_n 2^n \:a_{2^n}=\sum_{n\geq2} \frac{2^n}{2^n \cdot n \ln2\cdot(\ln n +\ln 2)^2}\leq \frac1{\ln 2}\sum_{n\geq2} \frac{1}{n \ln^2 n}.$$
The latter series may be treated in the same way, you may consider
$$
\sum_{n\geq2}\frac{2^n }{2^n \ln^2 (2^n)}= \frac1{\ln^2 2}\sum_{n\geq2}\frac{1}{n^2 }.
$$
Your initial series is convergent.
