# Determine whether $\sum_{n=1}^{\infty}\frac{(-1)^n}{n \log^2(n+1)}$ converges absolutely or conditionally. [duplicate]

Problem

Let $S = \sum_{n=1}^{\infty}\frac{(-1)^n}{n\log^2(n+1)}$.

Determine the series converges absolutely or conditionally.

Attempt

$S=\sum_{n=1}^{\infty}( -1)^n a_n$

$a_n$ is monotonically decreasing and it approaches zero when $n$ approaches infinity. So series is convergent .

Doubt

How to check for absolute convergence?

Ratio test fails here. Root test is of no use. I have attempted comparison tests using the fact that $n>\log(n)$, but no success there also.

## marked as duplicate by Fabian, Robert Z real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 30 '18 at 10:21

One option is the condensation test, which says $\sum_{n\geq 1}\frac{1}{n(\log(n+1))^2}$ converges if and only if $\sum_{n\geq 0}\frac{2^n}{2^n(\log (2^{n+1}))^2}$ does.