How to show that the series convergent


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

  • 2
    $\begingroup$ You have two different series there, But anyway, both compare to $\sum 1/n^2$. $\endgroup$ – 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$$\frac1{n^2\log n}\leqslant\frac1{n^2}.$$

  • 1
    $\begingroup$ I think you mean $\frac {1}{n^2}.$ Nonetheless, the point stands. $\endgroup$ – Doug M Jun 1 '18 at 23:14

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