Prove $\sum_{n=1}^{\infty}\arctan{\frac{1}{n^2+1}}=\arctan\left(\frac{\tan\pi\sqrt{(\sqrt{2}-1)/2}}{\tanh\pi\sqrt{(\sqrt{2}+1)/2}}\right)-\dfrac\pi8$

Question

My problem: Show that $$\sum_{n=1}^{\infty}\arctan{\frac{1}{n^2+1}}=\arctan\left(\frac{\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}}\right)}{\tanh\left(\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)}\right)-\frac\pi8$$

This result is nice because I know the famous problem $\sum_{n=1}^{\infty}\arctan{\frac{2}{n^2}}=\frac{3\pi}{4}$ and the following problem from the March $1990$ American Mathematical Monthly (Vol. $91$, P.$239$, Elementary Problem $3375$) by R.J. Chapman (Merton College, Oxford, U.K.):

Express $\sum_{n=1}^{\infty}\arctan{\frac{1}{n^2}}$ in closed form (in terms of a finite number of trigonometric and hyperbolic functions).

The following is the solution from the Aug-Sep $1991$ edition (Vol. 98, P.652-653) by Amites Sarkar, Winchester College, Hampshire, Great Britain: $$ \begin{aligned} \tan \left\{\sum_{n=1}^{\infty} \arctan \left(1 / n^2\right)\right\} & =\tan \left\{\sum_{n=1}^{\infty} \operatorname{Arg}\left(1+\frac{i}{n^2}\right)\right\} \\ & =\tan \left\{\operatorname{Arg}\left(\prod_{n=1}^{\infty}\left(1+\frac{(\pi(1+i) / \sqrt{2})^2}{n^2 \pi^2}\right)\right)\right\} \\ & =\tan \left\{\operatorname{Arg}\left(\frac{\sinh (\pi(1+i) / \sqrt{2})}{\pi(1+i) / \sqrt{2}}\right)\right\} \\ & =\tan \left\{\arctan \left(\frac{\tan (\pi / \sqrt{2})-\tanh (\pi / \sqrt{2})}{\tan (\pi / \sqrt{2})+\tanh (\pi / \sqrt{2})}\right)\right\} \\ & =\frac{\tan (\pi / \sqrt{2})-\tanh (\pi / \sqrt{2})}{\tan (\pi / \sqrt{2})+\tanh (\pi / \sqrt{2})} \end{aligned} $$

So we finally have

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

I tried to use these methods for my problem but I have failed. Thank you for your help.

This problem is similar to $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ and the harder problem $\sum_{n=1}^{\infty}\frac{1}{n^2+1}=\frac{1}{2}(\pi\coth{\pi}-1)$.

Answer 1

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.

Answer 2

We know that

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$\Gamma \left( 1+z \right)\Gamma \left( 1-z \right)=\frac{\pi z}{\sin \pi z}$

$\arctan \theta =\frac{i}{2}\ln \left( \frac{i+\theta }{i-\theta } \right)$

$$S=\sum\limits_{k=1}^{+\infty }{\arctan \frac{1}{1+k^{2}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\arctan \frac{1}{1+k^{2}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\frac{i}{2}\ln \left( \frac{i+\frac{1}{1+k^{2}}}{i-\frac{1}{1+k^{2}}} \right)}$$

$$=\frac{i}{2}\underset{n\to +\infty }{\mathop{\lim }}\,\ln \prod\limits_{k=1}^{n}{\left( \frac{i+\frac{1}{1+k^{2}}}{i-\frac{1}{1+k^{2}}} \right)}=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=1}^{n}{\left( \frac{i+\frac{1}{1+k^{2}}}{i-\frac{1}{1+k^{2}}} \right)} \right)$$

$$=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=1}^{n}{\left( \frac{i\left( 1+k^{2} \right)+1}{i\left( 1+k^{2} \right)-1} \right)} \right)=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=1}^{n}{\left( \frac{1+k^{2}-i}{1+k^{2}+i} \right)} \right)$$

$$=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=1}^{n}{\left( \frac{k^{2}+1-i}{k^{2}+1+i} \right)} \right)=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=0}^{n}{\left( \frac{\left( k+1 \right)^{2}+1-i}{\left( k+1 \right)^{2}+1+i} \right)} \right)$$

Note that $$\left( k+1 \right)^{2}+1\pm i=\left( k+1+i\sqrt{1\pm i} \right)\left( k+1-i\sqrt{1\pm i} \right)$$ $$=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\frac{\prod\limits_{k=0}^{n}{\left( k+1+i\sqrt{1-i} \right)}}{\prod\limits_{k=0}^{n}{\left( k+1+i\sqrt{1+i} \right)}}\cdot \frac{\prod\limits_{k=0}^{n}{\left( k+1-i\sqrt{1-i} \right)}}{\prod\limits_{k=0}^{n}{\left( k+1-i\sqrt{1+i} \right)}} \right)$$

$$=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\frac{\frac{1}{n!n^{1+i\sqrt{1-i}}}\prod\limits_{k=0}^{n}{\left( k+1+i\sqrt{1-i} \right)}}{\frac{1}{n!n^{1+i\sqrt{1+i}}}\prod\limits_{k=0}^{n}{\left( k+1+i\sqrt{1+i} \right)}}\cdot \frac{\frac{1}{n!n^{1-i\sqrt{1-i}}}\prod\limits_{k=0}^{n}{\left( k+1-i\sqrt{1-i} \right)}}{\frac{1}{n!n^{1-i\sqrt{1+i}}}\prod\limits_{k=0}^{n}{\left( k+1-i\sqrt{1+i} \right)}} \right)$$

$$=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\frac{\prod\limits_{k=0}^{n}{\left( k+1+i\sqrt{1-i} \right)}}{n!n^{1+i\sqrt{1-i}}}\cdot \frac{n!n^{1+i\sqrt{1+i}}}{\prod\limits_{k=0}^{n}{\left( k+1+i\sqrt{1+i} \right)}}\cdot \frac{\prod\limits_{k=0}^{n}{\left( k+1-i\sqrt{1-i} \right)}}{n!n^{1-i\sqrt{1-i}}}\cdot \frac{n!n^{1-i\sqrt{1+i}}}{\prod\limits_{k=0}^{n}{\left( k+1-i\sqrt{1+i} \right)}} \right)$$

$$=\frac{i}{2}\ln \left( \frac{\Gamma \left( 1+i\sqrt{1+i} \right)\cdot \Gamma \left( 1-i\sqrt{1+i} \right)}{\Gamma \left( 1+i\sqrt{1-i} \right)\cdot \Gamma \left( 1+i\sqrt{1-i} \right)} \right)$$

But $$i\sqrt{1\pm i}=i\sqrt{\sqrt{2}\left( \cos \frac{\pi }{4}\pm i\sin \frac{\pi }{4} \right)}=\sqrt[4]{2}i\exp \pm \frac{\pi }{8}=\sqrt[4]{2}i\left( \cos \pm \frac{\pi }{8}+i\sin \pm \frac{\pi }{8} \right)$$

$$=\sqrt[4]{2}i\left( \cos \frac{\pi }{8}\pm i\sin \frac{\pi }{8} \right)=\sqrt[4]{2}\left( \mp \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right)$$

Then $$=\frac{i}{2}\ln \left( \frac{\Gamma \left( 1+\sqrt[4]{2}\left( -\sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)\cdot \Gamma \left( 1-\sqrt[4]{2}\left( -\sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)}{\Gamma \left( 1+\sqrt[4]{2}\left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)\cdot \Gamma \left( 1-\sqrt[4]{2}\left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)} \right)$$

$$=\frac{i}{2}\ln \left( \frac{\pi \left( -\sqrt[4]{2}\sin \frac{\pi }{8}+\sqrt[4]{2}i\cos \frac{\pi }{8} \right)}{\sin \left( \pi \left( -\sqrt[4]{2}\sin \frac{\pi }{8}+\sqrt[4]{2}i\cos \frac{\pi }{8} \right) \right)}\cdot \frac{\sin \left( \pi \left( \sqrt[4]{2}\sin \frac{\pi }{8}+\sqrt[4]{2}i\cos \frac{\pi }{8} \right) \right)}{\pi \left( \sqrt[4]{2}\sin \frac{\pi }{8}+\sqrt[4]{2}i\cos \frac{\pi }{8} \right)} \right)$$

$$=\frac{i}{2}\ln \left( \frac{-\sqrt[4]{2}\sin \frac{\pi }{8}+\sqrt[4]{2}i\cos \frac{\pi }{8}}{\sqrt[4]{2}\sin \frac{\pi }{8}+\sqrt[4]{2}i\cos \frac{\pi }{8}}\cdot \frac{\sin \left( \sqrt[4]{2}\pi \left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)}{\sin \left( \sqrt[4]{2}\pi \left( -\sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)} \right)$$

Note that $$\frac{-\sqrt[4]{2}\sin \frac{\pi }{8}+\sqrt[4]{2}i\cos \frac{\pi }{8}}{\sqrt[4]{2}\sin \frac{\pi }{8}+\sqrt[4]{2}i\cos \frac{\pi }{8}}=\frac{-\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}=-\frac{\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}}{\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}\cdot \frac{\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}}{\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}}$$

$$=-\left( \sin ^{2}\frac{\pi }{8}-\cos ^{2}\frac{\pi }{8}-2i\sin \frac{\pi }{8}\cos \frac{\pi }{8} \right)=\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}=\exp \left( \frac{\pi i}{4} \right)$$

Then $$=\frac{i}{2}\ln \left( \exp \left( \frac{\pi i}{4} \right)\cdot \frac{\sin \left( \sqrt[4]{2}\pi \left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)}{\sin \left( \sqrt[4]{2}\pi \left( -\sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)} \right)$$

$$=\frac{i}{2}\ln \left( -\exp \left( \frac{\pi i}{4} \right)\cdot \frac{\sin \left( \sqrt[4]{2}\pi \left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)}{\sin \left( \sqrt[4]{2}\pi \left( \sin \frac{\pi }{8}-i\cos \frac{\pi }{8} \right) \right)} \right)$$

On the other hand, Working expression $$\frac{\sin \left( \sqrt[4]{2}\pi \left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)}{\sin \left( \sqrt[4]{2}\pi \left( \sin \frac{\pi }{8}-i\cos \frac{\pi }{8} \right) \right)}=\frac{e^{\sqrt[4]{2}\pi i\left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right)}-e^{-\sqrt[4]{2}\pi i\left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right)}}{e^{\sqrt[4]{2}\pi i\left( \sin \frac{\pi }{8}-i\cos \frac{\pi }{8} \right)}-e^{-\sqrt[4]{2}\pi i\left( \sin \frac{\pi }{8}-i\cos \frac{\pi }{8} \right)}}$$

$$=\frac{e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}}{e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}}}=\frac{2e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}}-2e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}}{2e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}-2e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}}}$$

$$=\frac{\left( e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}+e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}} \right)\left( e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}} \right)-\left( e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}+e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}} \right)\left( e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}} \right)}{\left( e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}+e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}} \right)\left( e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}} \right)+\left( e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}} \right)\left( e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}+e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}} \right)}$$

$$=\frac{-\frac{1}{i}\cdot \left( -i\frac{e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}}{e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}+e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}} \right)-\frac{e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}}}{e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}+e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}}}}{-\frac{1}{i}\cdot \left( -i\frac{e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}}{e^{\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}+e^{-\sqrt[4]{2}\pi i\sin \frac{\pi }{8}}} \right)+\frac{e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}-e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}}}{e^{\sqrt[4]{2}\pi \cos \frac{\pi }{8}}+e^{-\sqrt[4]{2}\pi \cos \frac{\pi }{8}}}}$$

$$=\frac{-\frac{1}{i}\tan \left( \sqrt[4]{2}\pi \sin \frac{\pi }{8} \right)-\tanh \left( \sqrt[4]{2}\pi \cos \frac{\pi }{8} \right)}{-\frac{1}{i}\tan \left( \sqrt[4]{2}\pi \sin \frac{\pi }{8} \right)+\tanh \left( \sqrt[4]{2}\pi \cos \frac{\pi }{8} \right)}=-\frac{i\tanh \left( \sqrt[4]{2}\pi \cos \frac{\pi }{8} \right)+\tan \left( \sqrt[4]{2}\pi \sin \frac{\pi }{8} \right)}{i\tanh \left( \sqrt[4]{2}\pi \cos \frac{\pi }{8} \right)-\tan \left( \sqrt[4]{2}\pi \sin \frac{\pi }{8} \right)}$$

$$=-\frac{i+\frac{\tan \left( \sqrt[4]{2}\pi \sin \frac{\pi }{8} \right)}{\tanh \left( \sqrt[4]{2}\pi \cos \frac{\pi }{8} \right)}}{i-\frac{\tan \left( \sqrt[4]{2}\pi \sin \frac{\pi }{8} \right)}{\tanh \left( \sqrt[4]{2}\pi \cos \frac{\pi }{8} \right)}}$$

Then the summ is equivalent to $$S=\frac{i}{2}\ln \left( \exp \left( \frac{\pi i}{4} \right)\cdot -\frac{\sin \left( \sqrt[4]{2}\pi \left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right) \right)}{\sin \left( \sqrt[4]{2}\pi \left( \sin \frac{\pi }{8}-i\cos \frac{\pi }{8} \right) \right)} \right)$$

$$=-\frac{\pi }{8}+\frac{i}{2}\ln \left( \frac{i+\frac{\tan \left( \sqrt[4]{2}\pi \sin \frac{\pi }{8} \right)}{\tanh \left( \sqrt[4]{2}\pi \cos \frac{\pi }{8} \right)}}{i-\frac{\tan \left( \sqrt[4]{2}\pi \sin \frac{\pi }{8} \right)}{\tanh \left( \sqrt[4]{2}\pi \cos \frac{\pi }{8} \right)}} \right)=\tan ^{-1}\left( \frac{\tan \left( \sqrt[4]{2}\pi \cos \frac{\pi }{8} \right)}{\tanh \left( \sqrt[4]{2}\pi \sin \frac{\pi }{8} \right)} \right)-\frac{\pi }{8}$$

$$=\tan ^{-1}\left( \frac{\tan \left( \sqrt[4]{2}\pi \frac{\sqrt{2+\sqrt{2}}}{2} \right)}{\tanh \left( \sqrt[4]{2}\pi \frac{\sqrt{2-\sqrt{2}}}{2} \right)} \right)-\frac{\pi }{8}=\tan ^{-1}\left( \frac{\tan \left( \pi \frac{\sqrt{\sqrt{2}+1}}{2} \right)}{\tanh \left( \pi \frac{\sqrt{\sqrt{2}-1}}{2} \right)} \right)-\frac{\pi }{8}$$ :D

Answer 3

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.