Show that the series $\frac{1}{2^2\log{2}}-\frac{1}{3^2\log{3}}+\frac{1}{4^2\log{4}}-\frac{1}{5^2\log{5}}+\ldots\,$ converges

How to show that the series convergent

$$\frac{1}{2^2\log{2}}-\frac{1}{3^2\log{3}}+\frac{1}{4^2\log{4}}-\dots$$

The series can be written as $$\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n+1)^{2}\log{(n+1)}}$$ I want to use Leibnitz's test. Here $u_n=\frac{1}{(n+1)^{2}\log{(n+1)}}\to 0$ as $n\to \infty$

How to show $u_n$ is monotone decreasing?

Is there any other method to solve except Leibnitz

• You have two different series there, But anyway, both compare to $\sum 1/n^2$. – Lord Shark the Unknown Jun 1 '18 at 7:14

This series not only converges, but it converges absolutely since $$\left|\frac{(-1)^{n+1}}{(n+1)^2\log (n+1)}\right|\le \frac{1}{(n+1)^2}$$ and $$\sum \frac{1}{(n+1)^2}$$ converges.
Why Leibniz? The series converges absolutely, since$$\frac1{n^2\log n}\leqslant\frac1{n^2}.$$
• I think you mean $\frac {1}{n^2}.$ Nonetheless, the point stands. – Doug M Jun 1 '18 at 23:14