"Summing" the series $\sin(x)+\frac{1}{2}\sin(2x)+\frac{1}{3}\sin(3x)+\frac{1}{4}\sin(4x)+...$ "Summing" the series $\sin(x)+\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)+\dfrac{1}{4}\sin(4x)+...$
Pose $$S=\sin(x)+\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)+\dfrac{1}{4}\sin(4x)+...$$
$$C=\cos(x)+\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)+\dfrac{1}{4}\cos(4x)+...$$
$$C+iS = e^{ix}+\dfrac{1}{2}(e^{ix})^2+\dfrac{1}{3}(e^{ix})^3+\dfrac{1}{4}(e^{ix})^4+...$$
Let $t$ = $e^{ix}$
Then we have a series of $t+\dfrac{t}{2}+\dfrac{t}{3}+\dfrac{t}{4}+...=\log(1+t)$
Which is $-\log(1-e^{ix})=\log(1-\cos(x)-i\sin(x))$, use the formula $\log(A+iB)=\dfrac{1}{2}\log(A^2+B^2)+\arctan\left(\dfrac{B}{A}\right)$
$-\log([1-\cos(x)]-i\sin(x))=-\dfrac{1}{2}\log([1-\cos(x)]^2-i\sin^2(x))-i\arctan\left(-\dfrac{\sin(x)}{1-\cos(x)}\right))$. Since we are interested only in the imaginary part, we have the sum for $S$ is:
$$-\arctan\left(-\dfrac{\sin(x)}{1-\cos(x)}\right)$$
Which is $-\arctan\left(-\cot(\dfrac{x}{2})\right)$
I don't know what to do next.
The "sum" of this series should be $\dfrac{\pi-x}{2}$, according to Euler.
 A: First, the series are
$$
\log(1+x)=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^k}k
$$
and
$$
-\log(1-x)=\sum_{k=1}^\infty\frac{x^k}k
$$
The series is $2\pi$-periodic and the sum for $-\pi\le x\le\pi$ is
$$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(kx)}k
&=\frac1{2i}\sum_{k=1}^\infty\frac{e^{ikx}-e^{-ikx}}k\\
&=-\frac1{2i}\left(\log(1-e^{ix})-\log(1-e^{-ix})\right)\\
&=\frac1{2i}\log\left(\frac{1-e^{-ix}}{1-e^{ix}}\right)\\
&=\frac1{2i}\log\left(-e^{-ix}\right)\\[3pt]
&=\operatorname{sgn}(x)\frac\pi2-\frac x2
\end{align}
$$

Note about $\boldsymbol{\operatorname{sgn}(x)}$
Since $\sin(x)$ is odd, the series should be odd. However, $\frac{\pi-x}2$ is not odd. The limit as $x\to0^+$ is $\frac\pi2$, but at $x=0$, all the terms of the series are $0$, so the sum is $0$. So that should clue us into the fact that there is a discontinuity there (this is because the convergence is not uniform near $0$).
Taking the oddness into account we get $\operatorname{sgn}(x)\frac\pi2-\frac x2$. This is also shown by looking at $\frac1{2i}\log\left(-e^{-ix}\right)$: when $x$ is increasing from $0$, $-e^{-ix}$ is moving clockwise from $-1$; thus, $\frac1{2i}\log\left(-e^{-ix}\right)$ is decreasing from $\frac\pi2$. When $x$ is decreasing from $0$, $-e^{-ix}$ is moving counter-clockwise from $-1$; thus, $\frac1{2i}\log\left(-e^{-ix}\right)$ is increasing from $-\frac\pi2$.
