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

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.

• Is the series $\sum_{n=1}^\infty\frac{(-1)^n}{n\log^2(n+1)}$? – José Carlos Santos Jul 30 '18 at 10:12
• To test the absolute convergence, you could try Cauchy's integral test. – xbh Jul 30 '18 at 10:15
• Heh. When I first read this question, I assumed that $\log^2$ was the iterated logarithm ($\log\circ\log$), not the square. I am not sure if there is any firmly established convention in this area. – Harald Hanche-Olsen Jul 30 '18 at 11:18

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

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

Yes, you are correct, ratio test fails here. Hint: note that for $n\geq 3$, $$0\leq \frac{1}{n(\log^2(n+1))}\leq \frac{1}{n(\log^2(n))}\leq \int_{n-1}^{n}\frac{dx}{x(\log^2(x))}=\frac{1}{\log(n-1)}-\frac{1}{\log(n)}.$$

What may we conclude?

Simply we have $$\sum_{n=2}^\infty\frac{1}{n\log^2(n+1)}<\sum_{n=2}^\infty\frac{1}{n\log^2(n)}$$ and one may use the integral test for evaluating $\displaystyle\int_{2}^\infty\frac{dx}{x\log^2x}=\dfrac12$.