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Question Number 27189 by abdo imad last updated on 02/Jan/18

$$letgive{S}_n=\sum _{p=1}^{p=n}arctan\left(\frac{1}{2p^2}\right)find{lim}_{n->\propto }{S}_n.$$

Answered by prakash jain last updated on 03/Jan/18

$${\tan }^{-1}\frac{1}{2p^2}={\tan }^{-1}\frac{(2p+1)-(2p-1)}{1+(2p+1)(2p+1)}$$ $$={\tan }^{-1}(2p+1)-{\tan }^{-1}(2p-1)$$ $$p:1\to {\tan }^{-1}3-{\tan }^{-1}1$$ $$p:2\to {\tan }^{-1}5-{\tan }^{-1}3$$ $$⋮$$ $$p:n-1\to {\tan }^{-1}(2n-1)-{\tan }^{-1}(2n-3)$$ $$p:n\to {\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-3>1)$$ $$\underset{p=1}{\sum ^n}{\tan }^{-1}\frac{1}{2p^2}={\tan }^{-1}(2n+1)-{\tan }^{-1}1$$ $$\underset{n\to \mathrm{\infty }}{\lim }\underset{p=1}{\sum ^n}{\tan }^{-1}\frac{1}{2p^2}=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$

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