Converges or diverges: $\sum_{n=1}^{\infty}\left [\arctan{\frac{(-1)^n}{n}}+{(-1)^n\over n^2}\right]$

Question

How can I show that the following series converges or diverges ?

$$\sum_{n=1}^{\infty}\left [\arctan{\frac{(-1)^n}{n}}+{(-1)^n\over n^2}\right]$$

$\sum_{n=1}^{\infty}\left [\arctan{\frac{(-1)^n}{n}}\right]+\sum_{n=1}^{\infty}\left[{(-1)^n\over n^2}\right]$, for $\sum_{n=1}^{\infty}\left[{(-1)^n\over n^2}\right]$ I used Leibniz criteria, by showing that $1\over n^2$ is decreasing and its limit is zero, thus the series converges. For $\arctan...$ the first series I thought to use the same Leibniz criteria, because $1\over n$ is decreasing to zero, but I do not know, is it right to say that if ${(-1)^n \over n}$ decreases then $\left [\arctan{\frac{(-1)^n}{n}}\right]$ also decreases ? Is it right what I did ? Thank you.

Answer 1

For small $z$, $\arctan z$ behaves exactly like $z$: $$\arctan z = z+O(z^3) $$ hence $\sum_{n\geq 1}\arctan\frac{(-1)^n}{n}$ converges since $\sum_{n\geq 1}\frac{(-1)^n}{n}$ converges by Leibniz' test. By the way, $$ \sum_{n=1}^{+\infty}\arctan\frac{(-1)^n}{n}=\operatorname{Arg}\frac{\Gamma\left(\frac{1+i}{2}\right)}{\Gamma\left(1+\frac{i}{2}\right)}=-0.506671\ldots$$

Answer 2

Hint. Recall that, for $x$ near $0$, $$ \arctan x=x+\mathcal{O}\left(x^3\right) $$ then $$ \sum\arctan{\frac{(-1)^n}{n}}=\sum\frac{(-1)^n}{n}+\sum\mathcal{O}\left(\frac{1}{n^3}\right) $$ is convergent being the sum of two convergent series.