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I would like to prove the following series: $$\sum_{n=1}^\infty {(-1)^n\cdot \arctan\left(\frac{n}{1+n^2}\right)} $$ is convergent (absolutely?) or divergent. I think $\arctan\left(\frac{n}{1+n^2}\right)$ is divergent, but I don't know how it interacts with $(-1)^n$ and how to prove it.

Any ideas would be greatly appreciated.

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    $\begingroup$ Doesn’t your arctan piece look very much like $1/n$? Shouldn’t it have the same convergence properties as $\sum(-1)^n/n$? $\endgroup$ – Lubin Dec 20 '15 at 13:44
  • $\begingroup$ As Lubin commented, there will be a problem without $(-1)^n$ but with it, it should converge. $\endgroup$ – Claude Leibovici Dec 20 '15 at 13:47
  • $\begingroup$ Try the Leibniz rule. $\endgroup$ – A.Γ. Dec 20 '15 at 13:52
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Your initial series is convergent. To prove it, you may just use the alternating series test:

  • the function $x \mapsto \arctan \left(\dfrac{x}{1+x^2}\right)$ is decreasing over $[1,\infty)$, since its derivative is negative over $[1,\infty)$: $\left(\arctan \left(\dfrac{x}{1+x^2}\right)\right)'= \dfrac{1-x^2}{1+3 x^2+x^4}\leq 0$,

and

  • as $x \to \infty$, you have $\arctan \left(\dfrac{x}{1+x^2}\right)\sim \dfrac1x \longrightarrow 0.$
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  • $\begingroup$ Is it possible to show it's decreasing without using derivations? Maybe $a_{n} - a_{n+1} >0$ ? And this test doesn't prove absolute convergence, no? $\endgroup$ – Mykybo Dec 20 '15 at 14:11
  • $\begingroup$ @Mykybo Yes, here you can also apply $\arctan a - \arctan b =\arctan \left( \frac{a-b}{1+ab}\right)$. $\endgroup$ – Olivier Oloa Dec 20 '15 at 14:17
  • $\begingroup$ (+1) @Mykybo It does not converge absolutely. It is curious that this series can be evaluated exactly in terms of gamma functions. $\endgroup$ – Start wearing purple Dec 20 '15 at 14:18
  • $\begingroup$ @Startwearingpurple and how do I prove that it does not converge absolutely? $\endgroup$ – Mykybo Dec 20 '15 at 16:51
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Leibniz's criterion for alternating series works here: $\dfrac n{1+n^2}$ decreases to $0$, and $\arctan x$ is continuous increasing, hence $\arctan \dfrac n{1+n^2}$ decreases (to $0$).

It is not absolutely convergent, because $$\frac n{1+n^2}=\frac1n\cdot\frac1{1+\cfrac1{n^2}}=\frac1n+o\Bigl(\frac1n\Bigr)$$ Now $\arctan u=u+o(u)$, so
$$\arctan\frac n{1+n^2}=\frac1n+o\Bigl(\frac1n\Bigr)\sim_\infty \frac1n,$$ which diverges.

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  • $\begingroup$ Shouldn't we prove that it is decreasing? If I understand it correctly, what we do is ignore $\arctan{}$ because it is a continuously increasing function, which means it behaves in a same way as if it wasn't there. Also what criteria are you using to prove its not absolutely convergent? Also when you got this at the end: ${\frac{1}{n}}$, by $\sim_\infty$ you mean "as series goes to infinity"? $\endgroup$ – Mykybo Dec 20 '15 at 15:09
  • $\begingroup$ No, I use the notion of equivalent sequences at infinity. This means the ratio of the sequences tends to $1$. A general result in asymptotic analysis is that two series with positive equivalent terms both converge or both diverge. Here the problem comes down to the harmonic series. $\endgroup$ – Bernard Dec 20 '15 at 15:19
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To show that the terms are decreasing:

$\arctan\left(\frac{n}{1+n^2}\right)- \arctan\left(\frac{n+1}{1+(n+1)^2}\right) =\arctan\left(\dfrac{\frac{n}{1+n^2}-\frac{n+1}{1+(n+1)^2}}{1+\frac{n}{1+n^2}\frac{n+1}{1+(n+1)^2}}\right) $

and $\dfrac{n}{1+n^2}-\dfrac{n+1}{1+(n+1)^2} =\dfrac{n(1+(n+1)^2)-(n+1)(1+n^2)}{(1+n^2)(1+(n+1)^2)} $

and

$\begin{array}\\ n(1+(n+1)^2)-(n+1)(1+n^2) &=n(n^2+2n+2)-(n^3+n^2+n+1)\\ &=n^3+2n^2+2n-(n^3+n^2+n+1)\\ &=n^2+n-1\\ &> 0\\ \end{array} $

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