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Question Number 192278 by manxsol last updated on 13/May/23

$$S=arctan\left(\frac{2}{1^2}\right)+artan\left(\frac{2}{2^2}\right)+........$$

Commented by Frix last updated on 14/May/23

$$\mathrm{As}\mathrm{I}\mathrm{posted}\mathrm{before}\left(\mathrm{but}\mathrm{it}\mathrm{was}\mathrm{deleted}\right):$$$$\mathrm{I}\mathrm{approximated}\mathrm{it}\mathrm{using}\mathrm{software}\mathrm{and}\mathrm{got}$$$$S\approx 2.3561\mathrm{which}\mathrm{I}\mathrm{think}\mathrm{is}\frac{3\pi }{4}$$

Commented by manxsol last updated on 14/May/23

Answered by witcher3 last updated on 14/May/23

$$\mathrm{S}=\sum _{\mathrm{n}⩾1}{\tan }^{-1}\left(\frac{2}{{\mathrm{n}}^2}\right)=\sum _{\mathrm{n}⩾1}{\tan }^{-1}\left(\frac{\mathrm{n}+1-(\mathrm{n}-1)}{1+(\mathrm{n}-1)(1+\mathrm{n})}\right)$$$$={\tan }^{-1}(\mathrm{n}+1)-{\tan }^{-1}(\mathrm{n}-1)$$$$\mathrm{S}=\underset{x\to \mathrm{\infty }}{\lim }\underset{\mathrm{n}=1}{\sum ^{\mathrm{x}}}\{{\tan }^{-1}(\mathrm{n}+1)-{\tan }^{-1}(\mathrm{n}-1)\}$$$$=\underset{x\to \mathrm{\infty }}{\lim }[{\tan }^{-1}(\mathrm{x}+1)+{\tan }^{-1}\left(\mathrm{x}\right)-{\tan }^{-1}\left(1\right)\}$$$$=\frac{\pi }{2}+\frac{\pi }{2}-\frac{\pi }{4}=\frac{3\pi }{4}$$

Commented by manxsol last updated on 14/May/23

$$thank,Mr.Witcher$$

Commented by Frix last updated on 14/May/23

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