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Question Number 2751 by prakash jain last updated on 26/Nov/15

$$\mathrm{Prove}\mathrm{that}$$$$\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{\sin i}{i}=\frac{\pi }{2}-\frac{1}{2}$$

Answered by prakash jain last updated on 27/Nov/15

$$\mathrm{See}\mathrm{question}2735\mathrm{fo}\mathrm{convergence}\mathrm{proof}.$$$$\mathrm{Question}\mathrm{used}iasindexusingnasindex$$$$asi=\sqrt{-1}isusedinanswer.$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\sin n}{n}$$$$=\mathrm{\Im }\left[\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{e^{in}}{n}\right]=\mathrm{\Im }\left(\ln (1-e^i)\right)=\arg (1-e^i)=\frac{\pi -1}{2}$$

Commented by prakash jain last updated on 27/Nov/15

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\sin nx}{n}=\frac{\pi -x}{2}$$

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