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Question Number 196325 by sniper237 last updated on 22/Aug/23

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}arg(n^2+n+1+i)=\pi /2$$

Answered by witcher3 last updated on 22/Aug/23

$$=\sum _{\mathrm{n}⩾0}{\tan }^{-1}\left(\frac{1}{1+\mathrm{n}+{\mathrm{n}}^2}\right)=\sum _{\mathrm{n}⩾0}{\tan }^{-1}\left(\frac{\mathrm{n}+1-\mathrm{n}}{1+(\mathrm{n}+1)\mathrm{n}}\right)$$$$=\sum _{\mathrm{n}⩾0}{\tan }^{-1}(\mathrm{n}+1)-{\tan }^{-1}\left(\mathrm{n}\right)$$$$=\underset{x\to \mathrm{\infty }}{\lim }\underset{\mathrm{n}=0}{\sum ^{\mathrm{x}}}{\tan }^{-1}(\mathrm{n}+1)-{\tan }^{-1}\left(\mathrm{n}\right)={\underset{x\to \mathrm{\infty }}{\lim tan}}^{-1}(1+\mathrm{x})=\frac{\pi }{2}$$

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