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Given the number series:

$\begin{aligned}\sum_{n=1}^{+\infty}\end{aligned} (-1)^n \log\left(\frac{2}{\pi}\,\arctan\sqrt{n}\right) $

since both the criterion of Leibniz, is the absolute convergence criteria are inconclusive, I do not know how to proceed.

Some idea?

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    The Leibniz (alternating series) criterion is conclusive, and we find that the series converges. – 2017-01-12
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    $$\log\left(\frac{2}{\pi}\arctan\sqrt{n}\right) = \log\left(1-\frac{2}{\pi}\arctan\frac{1}{\sqrt{n}}\right)\sim -\frac{2}{\pi\sqrt{n}}. $$ – 2017-01-12

1 Answers 1

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The series equals:

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

Where:

$$a_n = \log \left( \frac{\pi}{2\arctan \sqrt n} \right)$$

$a_n$ is nonnegative, decreasing and goes to $0$, so by the alternating series test, the series converges.