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

Question

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?

Answer 1

After a lot of struggle I found this:

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

$$S=\frac{1}{2}\left|-ln(1+ie^{-i})+ln(1+ie^i)\right|$$

Answer 2

Firstly, we add $C $and $iS$ together $$ \begin{aligned} C+iS&=\sum_{n=1}^{\infty}\left(\frac{e^i}{i}\right)^n \frac{1}{n}=-\ln \left(1-\frac{e^i}{i}\right) =-\ln \left(1+i e^i\right) \cdots(1) \end{aligned} $$ Secondly, we subtract $iS$ from $C$ $$ \begin{aligned} C- & i S=-\ln \left(1-\frac{e^{-i}}{i}\right)=-\ln \left(1+i e^{-i}\right) \cdots(2) \end{aligned} $$ $(1)-(2)$ yields $$ S=\frac{1}{2 i} \ln \left(\frac{1+i e^{-i}}{1+i e^i}\right) $$ $$|S|= \frac{1}{2}\left|\ln \left(\frac{1+i e^{-i}}{1+i e^i}\right)\right|$$

We are now going to express $\frac{1+i e^{-i}}{1+i e^i}$ in polar form.

$$ \begin{aligned} \frac{1+i e^{-i}}{1+i e^i}& =\frac{1+e^{\frac{\pi-2 }{2}i}}{1+e^{\frac{\pi+2}{2} i}} = \frac{e^{\frac{\pi-2}{4} i}\left(e^{\frac{\pi-2}{4} i}+e^{-\frac{\pi-2}{4} i}\right)}{e^{\frac{\pi+2}{4} i}\left(e^{\frac{\pi+2}{4} i}+e^{-\frac{\pi+2}{4} i}\right)} = e^{-i} \frac{\cos \left(\frac{\pi-2}{4}\right)}{\cos \left(\frac{\pi+2}{4}\right)} \end{aligned} $$

Then $$ \ln \left(\frac{1+i e^{-i}}{1+i e^i}\right) = \ln \left[\frac{\cos \left(\frac{\pi-2}{4}\right)}{\cos \left(\frac{\pi+2}{4}\right)}\right]-i $$ We can now conclude that $$ |S |=\frac{1}{2} \sqrt{\ln ^2\left[\frac{\cos \left(\frac{\pi-2}{4}\right)}{\cos \left(\frac{\pi+2}{4}\right)}\right]+1} $$