Question Number 2751 by prakash jain last updated on 26/Nov/15 | ||
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\:{i}}{{i}}=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ | ||
Answered by prakash jain last updated on 27/Nov/15 | ||
$$\mathrm{See}\:\mathrm{question}\:\mathrm{2735}\:\mathrm{fo}\:\mathrm{convergence}\:\mathrm{proof}. \\ $$$$\mathrm{Question}\:\mathrm{used}\:{i}\:{as}\:{index}\:{using}\:{n}\:{as}\:{index}\: \\ $$$${as}\:{i}=\sqrt{−\mathrm{1}}\:{is}\:{used}\:{in}\:{answer}. \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\:{n}}{{n}} \\ $$$$=\Im\left[\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{in}} }{{n}}\right]=\Im\left(\mathrm{ln}\:\left(\mathrm{1}−{e}^{{i}} \right)\right)=\mathrm{arg}\left(\mathrm{1}−{e}^{{i}} \right)=\frac{\pi−\mathrm{1}}{\mathrm{2}} \\ $$ | ||
Commented by prakash jain last updated on 27/Nov/15 | ||
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\:{nx}}{{n}}=\frac{\pi−{x}}{\mathrm{2}} \\ $$ | ||