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show that: $$\sum_{n=1}^{\infty}\arctan{\left(\dfrac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)\cdot\dfrac{e^{\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}+e^{-\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}}{e^{\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}-e^{-\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}}\right)}-\dfrac{\pi}{8}$$

This relust is nice.(maybe have some wrong)

becasue I know this famous problem $$\sum_{n=1}^{\infty}\arctan{\dfrac{2}{n^2}}=\dfrac{3\pi}{4}$$ and follow AMM( E3375) problem $$\sum_{n=1}^{\infty}\arctan{\dfrac{1}{n^2}}=\arctan{\left(\dfrac{\tan{\frac{\pi}{\sqrt{2}}}-th{\dfrac{\pi}{\sqrt{2}}}}{\tan{\dfrac{\pi}{\sqrt{2}}}+th{\dfrac{\pi}{\sqrt{2}}}}\right)}$$

Follow is AMM solution:enter image description here

My try: my problem I want use this methods,But at last failure it. Thank you for you help.

This problem is similar this:we konw this

$$\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$$ and then little hard problem: $$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{1}{2}(\pi\coth{\pi}-1)$$

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  • $\begingroup$ What is $th\frac{\pi}{\sqrt2}$? $\endgroup$ – mathlove Dec 20 '13 at 5:59
  • $\begingroup$ Hello,is $\tanh{\dfrac{\pi}{\sqrt{2}}}$ $\endgroup$ – math110 Dec 20 '13 at 6:02
  • $\begingroup$ The solution should be very similar because $$\prod_{n=1}^\infty \left(1 + \frac{i}{n^2+1}\right) = \prod_{n=1}^\infty \left(\frac{1 + \frac{1+i}{n^2}}{1+\frac{1}{n^2}}\right)$$ $\endgroup$ – achille hui Dec 20 '13 at 6:34
  • $\begingroup$ Does AMM stand for American Math Monthly? if not, what is it? $\endgroup$ – user2312512851 Apr 5 '18 at 18:29
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For any $\alpha > 0$, let $u + iv = \pi\sqrt{\alpha^2 + i}$, we have $$\begin{align} & \tan\left\{\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{n^2+\alpha^2}\right)\right\} =\tan\left\{\sum_{n=1}^\infty\arg\left(1+\frac{i}{n^2+\alpha^2}\right)\right\}\\ = &\tan\left\{\arg\prod_{n=1}^\infty \left(1+\frac{i}{n^2+\alpha^2}\right)\right\} =\tan\left\{\arg\prod_{n=1}^\infty \left( \frac{1+\frac{\alpha^2+i}{n^2}}{1+\frac{\alpha^2}{n^2}} \right)\right\}\\ = &\tan\left\{\arg\prod_{n=1}^\infty \left(1+\frac{\alpha^2+i}{n^2}\right)\right\} =\tan\left\{\arg\prod_{n=1}^\infty \left(1+\frac{(u+iv)^2}{n^2\pi^2}\right)\right\}\\ = & \tan\left\{\arg\left(\frac{\sinh(u+iv)}{u+iv}\right)\right\} =\tan\left\{\arg\left(\frac{\tanh u + i\tan v}{u+iv}\right)\right\}\\ = & \tan\left\{\arg(\tanh u + i\tan v) - \arg(u+iv)\right\}\\ = & \tan\left\{\tan^{-1}\left(\frac{\tan v}{\tanh u}\right)-\frac12\arg(\alpha^2 + i)\right\}\\ = & \tan\left\{\tan^{-1}\left(\frac{\tan v}{\tanh u}\right)-\frac12\tan^{-1}\left(\frac{1}{\alpha^2}\right)\right\} \end{align}$$ For $\alpha = 1$, we have $u + iv = \pi\sqrt{\frac{\sqrt{2}+1}{2}} + i\pi\sqrt{\frac{\sqrt{2}-1}{2}}$, so

$$\sum_{n=1}^\infty \tan^{-1}\left(\frac{1}{n^2+1}\right) = \tan^{-1}\left(\frac{\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}}\right)}{\tanh\left(\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)}\right)- \frac{\pi}{8} + N \pi$$

for some integer $N$ to be determined. Numerically, the RHS excluding the unknown term $N\pi$ is about $1.0373$. On the LHS, we know it is a number around $1$. So the unknown constant $N$ is $0$ and we are done.

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  • $\begingroup$ Thank you,It;s very nice!+1 $\endgroup$ – math110 Dec 22 '13 at 8:28
  • $\begingroup$ Could you explain the reason for taking the $\tan$ of the stuff inside of the brackets? It seems superfluous. $\endgroup$ – Random Variable Apr 8 '14 at 21:37
  • $\begingroup$ @RandomVariable 1) I'm following the steps of the AMM solution in question. 2) If I don't take $\tan$, then I need to keep track of the $N\pi$ factor in each step which is even more inelegant. $\endgroup$ – achille hui Apr 9 '14 at 4:49
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Here is a closed form

$$ -\frac{1}{4}-\frac{1}{4}\,{\frac {\pi \,\cot \left( \pi \,\sqrt {-1+i} \right) }{\sqrt {-1 +i}}}-\frac{1}{4} \,{\frac {\pi \,\cot \left( \pi \,\sqrt {-1-i} \right) }{ \sqrt {-1-i}}}\sim 0.9676963204 .$$

Maybe one can simplify further.

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