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Question Number 26812 by amit96 last updated on 30/Dec/17

$$sumofinfiniteseris$$$${\tan }^{-1}\left(2/n^2\right)$$

Answered by prakash jain last updated on 31/Dec/17

$$\frac{2}{n^2}=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}$$$${\tan }^{-1}\frac{2}{n^2}={\tan }^{-1}(n+1)-{\tan }^{-1}(n-1)$$$$1:{\tan }^{-1}2-{\tan }^{-1}0$$$$2:{\tan }^{-1}3-{\tan }^{-1}1$$$$3:{\tan }^{-1}4-{\tan }^{-1}2$$$$..$$$$(n-1):{\tan }^{-1}n-{\tan }^{-1}(n-2)$$$$n:{\tan }^{-1}(n+1)-{\tan }^{-1}(n-1)$$$$\mathrm{sum}\mathrm{of}n\mathrm{terms}$$$${\tan }^{-1}(n+1)+{\tan }^{-1}\left(n\right)-{\tan }^{-1}1-{\tan }^{-1}0$$$$\underset{j=1}{\sum ^{\mathrm{\infty }}}{\tan }^{-1}\frac{2}{n^2}=\frac{\pi }{2}+\frac{\pi }{2}-\frac{\pi }{4}=\frac{3\pi }{4}$$

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