Find-n-1-tan-1-1-n-2-

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

Find : \underset{n = 1}{\sum^{\infty}} \tan^{- 1} (\frac{1}{n^{2}})

Answered by mindispower last updated on 27/Jan/22

S = Σ \tan^{- 1} (\frac{1}{n^{2}}) = \sum_{n ⩾ 1} a r g (n^{2} + i) = a r g \prod_{n ⩾ 1} (1 + \frac{i}{n^{2}})

\frac{s i n (π x)}{π x} = \prod_{n ⩾ 1} (1 - \frac{x^{2}}{n^{2}})

x = e^{\frac{i 3 π}{4}}

\frac{s i n (- \frac{π}{\sqrt{2}} + \frac{i π}{\sqrt{2}})}{π e^{\frac{3 i π}{4}}} = \prod_{n ⩾ 1} (1 + \frac{i}{n^{2}})

= \frac{s i n (- \frac{π}{\sqrt{2}}) c o s (\frac{i π}{\sqrt{3}}) + c o s (\frac{π}{\sqrt{2}}) s i n (\frac{i π}{\sqrt{2}})}{π e^{\frac{3 i π}{4}}}

= \frac{e^{- \frac{3 i π}{4}}}{π} (s i n (\frac{π}{\sqrt{2}}) c h (\frac{π}{\sqrt{2}}) + i s h (\frac{π}{\sqrt{2}}) c o s (\frac{π}{\sqrt{2}}))

= - \frac{1}{π \sqrt{2}} (s i n (\frac{π}{\sqrt{2}}) c h (\frac{π}{\sqrt{2}}) - s h (\frac{π}{\sqrt{2}}) c o s (\frac{π}{\sqrt{2}}) + i (s i n (\frac{π}{\sqrt{2}}) c h (\frac{π}{\sqrt{2}}) + s h (\frac{π}{\sqrt{2}}) c o s (\frac{π}{\sqrt{2}}))

S = a r g {- \frac{1}{π \sqrt{2}} (s i n (\frac{π}{\sqrt{2}}) c h (\frac{π}{\sqrt{2}}) - s h (\frac{π}{\sqrt{2}}) c o s (\frac{π}{\sqrt{2}}) + i (s i n (\frac{π}{\sqrt{2}}) c h (\frac{π}{\sqrt{2}}) + s h (\frac{π}{\sqrt{2}}) c o s (\frac{π}{\sqrt{2}}))}

S = \tan^{- 1} (\frac{1 + t h (\frac{π}{\sqrt{2}}) c o t (\frac{π}{\sqrt{2}})}{1 - t h (\frac{π}{\sqrt{2}}) c o t (\frac{π}{\sqrt{2}})})

\sum_{n ⩾ 1} \tan^{- 1} (\frac{1}{n^{2}}) = \tan^{- 1} (\frac{1 + t h (\frac{π}{\sqrt{2}}) c o t (\frac{π}{\sqrt{2}})}{1 - t h (\frac{π}{\sqrt{2}}) c o t (\frac{π}{\sqrt{2}})})

Commented by HongKing last updated on 28/Jan/22

very nice solution thank you dear Sir

very nice solution thank you dear Sir

Commented by mindispower last updated on 28/Jan/22

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