Is $\sum_{n=1}^\infty\frac{(-1)^n}{n\log^2(n+1)}$ absolutely convergent

calculusreal-analysissequences-and-series

Consider the series
$$\sum_{n=1}^\infty\frac{(-1)^n}{n\log^2(n+1)}.$$
Determine whether it converges absolutely or conditionally.

My attempt

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

an 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.

Best Answer

For the absolute convergence by cauchy condensation test we can consider the convergenge of the condensed series $\sum 2^n a_{2^n}$ that is

$$\sum \frac{2^n}{2^n(\log^2(2^n+1))}=\sum \frac{1}{\log^2(2^n+1)}$$

which converges by limit comparison test with $\sum \frac1{n^2}$ indeed

$$\frac{1}{\log^2(2^n+1)}\sim\frac1{n^2\log^2 2}$$