Solve $ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$ Solve $$ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$$
I first tried to use Eulers formula
$$ \frac{1}{2i} \sum_{k = 1}^{ \infty} \frac{1}{k} \left( e^{2ik} - e^{-2ik}  \right)$$
However to use the geometric formula here, I must subtract the $k=0$ term and that term is undefinted  since $1/k$. I also end up with something that diverges in my calculations and since $\sin 2k$ is limited the serie should not diverge.
 A: The summation is as follows:
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{\sin(2an)}{n} = \frac{1}{2i} \, \sum_{n=1}^{\infty} \frac{ e^{2ai n} - e^{- 2ai n}}{n} \\
&= - \frac{1}{2i} \left( \ln(1 - e^{2ai}) - \ln(1 - e^{-2ai}) \right) \\
&= - \frac{1}{2i} \ln\left( \frac{1 - e^{2ai}}{1 - e^{- 2ai}} \right) = - \frac{1}{2i} \ln\left( - \frac{e^{ai}}{e^{-ai}} \cdot \frac{\sin(a)}{\sin(a)} \right) \\
&= - \frac{1}{2i} \ln\left( - e^{2ai} \right) = - \frac{1}{2i} \left( \ln(e^{- \pi i}) + \ln(e^{2ai}) \right) \\
&= - \frac{1}{2i} \left( - \pi i + 2ai \right) = \frac{ \pi - 2a}{2}.
\end{align}
This provides
\begin{align}
\sum_{n=1}^{\infty} \frac{\sin(2an)}{n} = \frac{ \pi - 2a}{2}.
\end{align}
A: For finite non-zero $a,$
$$\sum_{k=1}^\infty\frac{e^{2iak}}k=-\ln(1-e^{2ia})$$
$$1-e^{2ia}=-e^{ia}(e^{ia}-e^{-ia})=-e^{ia}[2i\sin(a)]$$
$$\ln(1-e^{2ia})=\ln(e^{i(a+2m\pi)})+\ln2+\ln(-i)+\ln[\sin(a)]$$
$$=i(a+2m\pi)+\ln2-\frac{i\pi}2+\ln[\sin(a)]$$  
as $-i=\cos\left(\frac\pi2\right)+i\sin\left(\frac\pi2\right)=e^{i\dfrac\pi2}$ and where $m$ is any integer
Hope you can take it home from here setting $a=1,-1$ one by one
