Prove-that-i-1-sin-i-i-pi-2-1-2-

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

Prove that

\underset{i = 1}{\sum^{\infty}} \frac{\sin i}{i} = \frac{π}{2} - \frac{1}{2}

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

See question 2735 fo convergence proof .

Question used i a s i n d e x u s i n g n a s i n d e x

a s i = \sqrt{- 1} i s u s e d i n a n s w e r .

\underset{n = 1}{\sum^{\infty}} \frac{\sin n}{n}

= ℑ [\underset{n = 1}{\sum^{\infty}} \frac{e^{i n}}{n}] = ℑ (\ln (1 - e^{i})) = \arg (1 - e^{i}) = \frac{π - 1}{2}

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

\underset{n = 1}{\sum^{\infty}} \frac{\sin n x}{n} = \frac{π - x}{2}