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Question Number 165215 by HongKing last updated on 27/Jan/22

$$\mathrm{Find}:\underset{\mathbf{n}=1}{\sum ^{\mathrm{\infty }}}{\tan }^{-1}\left(\frac{1}{{\mathrm{n}}^2}\right)$$

Answered by mindispower last updated on 27/Jan/22

$$S=\mathrm{\Sigma }{\tan }^{-1}\left(\frac{1}{n^2}\right)=\sum _{n⩾1}arg(n^2+i)=arg\prod _{n⩾1}(1+\frac{i}{n^2})$$$$\frac{sin\left(\pi x\right)}{\pi x}=\prod _{n⩾1}(1-\frac{x^2}{n^2})$$$$x=e^{\frac{i3\pi }{4}}$$$$\frac{sin(-\frac{\pi }{\sqrt{2}}+\frac{i\pi }{\sqrt{2}})}{\pi e^{\frac{3i\pi }{4}}}=\prod _{n⩾1}(1+\frac{i}{n^2})$$$$=\frac{sin(-\frac{\pi }{\sqrt{2}})cos\left(\frac{i\pi }{\sqrt{3}}\right)+cos\left(\frac{\pi }{\sqrt{2}}\right)sin\left(\frac{i\pi }{\sqrt{2}}\right)}{\pi e^{\frac{3i\pi }{4}}}$$$$=\frac{e^{-\frac{3i\pi }{4}}}{\pi }(sin\left(\frac{\pi }{\sqrt{2}}\right)ch\left(\frac{\pi }{\sqrt{2}}\right)+ish\left(\frac{\pi }{\sqrt{2}}\right)cos\left(\frac{\pi }{\sqrt{2}}\right))$$$$=-\frac{1}{\pi \sqrt{2}}(sin\left(\frac{\pi }{\sqrt{2}}\right)ch\left(\frac{\pi }{\sqrt{2}}\right)-sh\left(\frac{\pi }{\sqrt{2}}\right)cos\left(\frac{\pi }{\sqrt{2}}\right)+i(sin\left(\frac{\pi }{\sqrt{2}}\right)ch\left(\frac{\pi }{\sqrt{2}}\right)+sh\left(\frac{\pi }{\sqrt{2}}\right)cos\left(\frac{\pi }{\sqrt{2}}\right))$$$$S=arg\{-\frac{1}{\pi \sqrt{2}}(sin\left(\frac{\pi }{\sqrt{2}}\right)ch\left(\frac{\pi }{\sqrt{2}}\right)-sh\left(\frac{\pi }{\sqrt{2}}\right)cos\left(\frac{\pi }{\sqrt{2}}\right)+i(sin\left(\frac{\pi }{\sqrt{2}}\right)ch\left(\frac{\pi }{\sqrt{2}}\right)+sh\left(\frac{\pi }{\sqrt{2}}\right)cos\left(\frac{\pi }{\sqrt{2}}\right))\}$$$$$$$$$$$$S={\tan }^{-1}\left(\frac{1+th\left(\frac{\pi }{\sqrt{2}}\right)cot\left(\frac{\pi }{\sqrt{2}}\right)}{1-th\left(\frac{\pi }{\sqrt{2}}\right)cot\left(\frac{\pi }{\sqrt{2}}\right)}\right)$$$$\sum _{n⩾1}{\tan }^{-1}\left(\frac{1}{n^2}\right)={\tan }^{-1}\left(\frac{1+th\left(\frac{\pi }{\sqrt{2}}\right)cot\left(\frac{\pi }{\sqrt{2}}\right)}{1-th\left(\frac{\pi }{\sqrt{2}}\right)cot\left(\frac{\pi }{\sqrt{2}}\right)}\right)$$

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

$$WithePleasurhaveaniceday$$

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