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Question Number 104016 by Dwaipayan Shikari last updated on 18/Jul/20

Answered by OlafThorendsen last updated on 18/Jul/20

$$\arctan x-\arctan y=\arctan \frac{x-y}{1+xy}$$$$\arctan (2k+1)-\arctan (2k-1)=\arctan \frac{(2k+1)-(2k-1)}{1+(2k+1)(2k-1)}$$$$=\arctan \frac{2}{4k^2}=\arctan \frac{1}{2k^2}$$$${\mathrm{S}}_n=\underset{k=1}{\sum ^{k=n}}\arctan \frac{1}{2k^2}$$$${\mathrm{S}}_n=\underset{k=1}{\sum ^{k=n}}\arctan (2k+1)-\arctan (2k-1)$$$${\mathrm{S}}_n=\arctan 3-\arctan 1$$$$+\arctan 5-\arctan 3...$$$$+\arctan (2k+1)-\arctan (2k-1)$$$${\mathrm{S}}_n=\arctan (2k+1)-\arctan 1$$$$\text{Double subscripts: use braces to clarify}$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\arctan \frac{1}{2k^2}=\frac{\pi }{4}$$$$$$

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