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Question Number 119969 by huotpat last updated on 28/Oct/20

Answered by Dwaipayan Shikari last updated on 28/Oct/20

$$\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}{tan}^{-1}\frac{1}{k^2+k+1}$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}{tan}^{-1}\frac{(k+1)-k}{1+k(k+1)}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}{tan}^{-1}(k+1)-{tan}^{-1}\left(k\right)$$$$={tan}^{-1}2-{tan}^{-1}1+{tan}^{-1}3-{tan}^{-1}2+...+{tan}^{-1}(n+1)-{tan}^{-1}\left(n\right)$$$$=\underset{n\to \mathrm{\infty }}{\lim }({tan}^{-1}(n+1)-{tan}^{-1}1)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$

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