# Show that the series converges absolutely $\sum (-1)^n \frac{1}{n (\log n)^2}$

Show that the series converges absolutely $$\sum (-1)^n \frac{1}{n (\log n)^2}$$

A series $\sum a_n$ converges absolutely if $\sum|a_n|< \infty$

Considering $a_n = (-1)^n \frac{1}{n (\log n)^2}$ such that $n>1$ $$|a_n| = \frac{1}{n( \log n)^2}$$

I am thinking here to use the comparison test to determine the convergence. How could I determine an expression that bound above and (eventually) below ? is there a more efficient test for this?

Also, by using the alternating series test, $\sum a_n$ converges. Can I use this fact, to determine the absolute convergence?

The alternating series test cannot be used to determine whether a series converges absolutely. To show that $\sum_n\frac{1}{n\log^2n}$ converges, I suggest using the integral test, since the integral $$\int_2^{\infty}\frac{dx}{x\log^2(x)}$$ can be evaluated by setting $u=\log x$.
Alternatively, one can use the Cauchy condensation test: since $(n\log^2n)^{-1}$ is decreasing to zero, we have
$$\sum_n\frac1{n\log^2n}<\infty\qquad\Leftrightarrow\qquad\sum_n2^n\frac1{2^n\log^22^n}<\infty;$$ the latter series is equal to $(\log2)^{-2}\sum_nn^{-2}$ which we know converges.