Summate using Abel method series $ \sum_{n=1}^{\infty} \sin{\theta n} $ Summate using Abel method series $ \sum_{n=1}^{\infty} \sin{\theta n} $.
For Abel sum we should consider functional series $ \sum_{n=1}^{\infty} \sin{\theta n} \cdot x^{n-1}$, find the sum $ S(x) $ and count $ \lim_{x \to 1-0} S(x) $.
I know that if there is a sum of Cesaro, sum of Abel also exists and equals to the sum of Cesaro. I obtained that sum of Cesaro (if I didn't mistake) is $ \cot{\frac{\theta}{2}} / 2 $. Hence the Abel sum $ \cot{\frac{\theta}{2}} / 2 $ too. But how to obtain Abel sum without using this theorem.
Thanks for the help!
 A: For $|x|<1$, we have
$$\begin{align}
\sum_{n=1}^\infty x^{n-1}\sin n\theta&=\frac1x\text{Im}\left(\sum_{n=1}^\infty x^{n}e^{in\theta}\right)\\\\
&=\text{Im}\left(\frac{e^{i\theta}}{1-xe^{i\theta}}\right)\\\\
&=\frac{\sin \theta}{x^2-2x\cos \theta+1}
\end{align}$$
Thus, 
$$\begin{align}
\lim_{x\to 1^-}\left(\sum_{n=1}^\infty x^{n-1}\sin n\theta\right)&=\frac{\sin \theta}{2-2\cos \theta}\\\\
&=\frac{\sin(2\theta/2)}{4\sin^2(\theta/2)}\\\\
&=\frac{2\sin (\theta/2)\cos(\theta/2)}{4\sin^2(\theta/2)}\\\\
&=\frac12 \cot(\theta/2)
\end{align}$$
as was to be shown!
A: By Euler's formulas: $\sin n\theta = \dfrac{e^{in\theta}-e^{-in\theta}}{2i}$, so
\begin{align}
\sum_{n=1}^\infty \sin(n\theta)\,x^{n-1} &=
\frac{x^{-1}}{2i} \sum_{n=1}^\infty \Big( (xe^{i\theta})^n- (xe^{-i\theta})^n \Big) \\
&= \frac{1}{2i} \Big( \frac{e^{i\theta}}{1-xe^{i\theta}} - \frac{e^{-i\theta}}{1-xe^{-i\theta}} \Big) 
\end{align}
(for $|x| < 1$) using the sum of two geometric series. Letting $x \to 1^-$, we get
\begin{align}
\frac{1}{2i} \Big( \frac{e^{i\theta}}{1-e^{i\theta}} - \frac{e^{-i\theta}}{1-e^{-i\theta}} \Big) &=
\frac{1}{2i} \Big( \frac{e^{i\theta}}{1-e^{i\theta}} - \frac{1}{e^{i\theta}-1} \Big) \\
&= \frac{1}{2i} \, \frac{1+e^{i\theta}}{1-e^{i\theta}} \\
&= \frac{i}{2} \, \frac{e^{i\theta/2}+e^{-i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}} = \frac12 \cot \frac{\theta}{2}.
\end{align}
