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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.

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    $\begingroup$ Note that $\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}$, what do you can conclude? $\endgroup$ – DiegoMath Jan 20 '15 at 19:12
  • $\begingroup$ That its increasing for any x ? $\endgroup$ – Ivan Gandacov Jan 20 '15 at 19:13
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    $\begingroup$ Leibniz criteria requires $a_n > 0$ and $a_n \to 0$ for the convergence of $\sum\limits_{n=1}^{\infty} (-1)^n a_n$, so yes it converges, for $a_n = \tan^{-1} \frac{1}{n}$ as it $\to 0$ as $n \to \infty$. $\endgroup$ – sciona Jan 20 '15 at 19:13
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    $\begingroup$ @JohnG. yes it does. $\endgroup$ – sciona Jan 20 '15 at 19:15
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    $\begingroup$ @JohnG. no it depends on $a_n$, the fact that $\frac{1}{n} \to 0$ and $\tan^{-1} \frac{1}{n} \to 0$ as well leads to $\sum\limits_{n=1}^{\infty} (-1)^n \tan^{-1} \frac{1}{n}$ to converge by Liebniz Criteria in this case. $\endgroup$ – sciona Jan 20 '15 at 19:21
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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, $$ \sum_{n=1}^{+\infty}\arctan\frac{(-1)^n}{n}=\operatorname{Arg}\frac{\Gamma\left(\frac{1+i}{2}\right)}{\Gamma\left(1+\frac{i}{2}\right)}=-0.506671\ldots$$

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    $\begingroup$ Dear God artillery to kill flies! $\endgroup$ – sciona Jan 20 '15 at 19:22
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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.

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