# Evaluate $S=\left|\sum_{n=1}^{\infty} \frac{\sin n}{i^n \cdot n}\right|$

Evaluate $$S=\left|\sum_{n=1}^{\infty} \dfrac{\sin n}{i^n \cdot n}\right|$$
where $i=\sqrt{-1}$

For this question, I did the following,

Let \begin{align*} S &= \sum_{n=1}^{\infty} \dfrac{\sin n}{i^n \cdot n} \\ C &= \sum_{n=1}^{\infty} \dfrac{\cos n}{i^n \cdot n} \end{align*} We have to evaluate $|S|$ $$\implies S = \Im{(C+iS)}=\Im\displaystyle \sum_{n=1}^{\infty} \dfrac{e^{in}}{i^n \cdot n}  However, due to the n in the denominator, I cannot sum the series. If only it had been in the numerator I would've sum it as an A.G.P. Can anyone suggest something? • aDo you know series for \ln(1+z)? – Robert Israel Jan 25 '15 at 19:00 • See here. – Mhenni Benghorbal Jan 25 '15 at 20:36 ## 1 Answer After a lot of struggle I found this:$$ S=\left|\sum_{n=1}^{\infty} \dfrac{\sin n}{i^n \cdot n}\right|=S=\frac{1}{2}\left|-ln(1+ie^{-i})+ln(1+ie^i)\right|

• Can you please provide the details too? Thanks! – user208998 Apr 26 '15 at 17:18
• I was struggling with this in school and solved it together with my prof. You've write the sum easier – Jan Apr 26 '15 at 17:20
• But can you at least provide some hints? – user208998 Apr 26 '15 at 17:23
• Use De Moivre and euler – Jan Apr 26 '15 at 17:25