Question Number 165215 by HongKing last updated on 27/Jan/22
$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right) \\ $$
Answered by mindispower last updated on 27/Jan/22
$${S}=\Sigma\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}{arg}\left({n}^{\mathrm{2}} +{i}\right)={arg}\underset{{n}\geqslant\mathrm{1}} {\prod}\left(\mathrm{1}+\frac{{i}}{{n}^{\mathrm{2}} }\right) \\ $$$$\frac{{sin}\left(\pi{x}\right)}{\pi{x}}=\underset{{n}\geqslant\mathrm{1}} {\prod}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$${x}={e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \\ $$$$\frac{{sin}\left(−\frac{\pi}{\:\sqrt{\mathrm{2}}}+\frac{{i}\pi}{\:\sqrt{\mathrm{2}}}\right)}{\pi{e}^{\frac{\mathrm{3}{i}\pi}{\mathrm{4}}} }=\underset{{n}\geqslant\mathrm{1}} {\prod}\left(\mathrm{1}+\frac{{i}}{{n}^{\mathrm{2}} }\right) \\ $$$$=\frac{{sin}\left(−\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cos}\left(\frac{{i}\pi}{\:\sqrt{\mathrm{3}}}\right)+{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){sin}\left(\frac{{i}\pi}{\:\sqrt{\mathrm{2}}}\right)}{\pi{e}^{\frac{\mathrm{3}{i}\pi}{\mathrm{4}}} } \\ $$$$=\frac{{e}^{−\frac{\mathrm{3}{i}\pi}{\mathrm{4}}} }{\pi}\left({sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){ch}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+{ish}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\pi\sqrt{\mathrm{2}}}\left({sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){ch}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)−{sh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+{i}\left({sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){ch}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+{sh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\right)\right. \\ $$$${S}={arg}\left\{−\frac{\mathrm{1}}{\pi\sqrt{\mathrm{2}}}\left({sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){ch}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)−{sh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+{i}\left({sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){ch}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+{sh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\right)\right\}\right. \\ $$$$ \\ $$$$ \\ $$$${S}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{th}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cot}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}{\mathrm{1}−{th}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cot}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}\right) \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{th}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cot}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}{\mathrm{1}−{th}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right){cot}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}\right) \\ $$
Commented by HongKing last updated on 28/Jan/22
$$\mathrm{very}\:\mathrm{nice}\:\mathrm{solution}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Sir} \\ $$
Commented by mindispower last updated on 28/Jan/22
$${Withe}\:{Pleasur}\:{have}\:{a}\:{nice}\:{day} \\ $$