Cesàro sum of the series $\sin x + \sin 2x + \sin 3x + \ldots = \frac{1}{2}\cot\frac{x}{2}$ for $x \neq 2k\pi, k \in \mathbb{Z}$ I'm learning about Fourier series (specifically Cesàro summation) and need help with the following problem:

Show that the Cesàro sum of the series $\sin x + \sin 2x + \sin 3x + \ldots$ is equal to $\frac{1}{2}\cot\frac{x}{2}$ for $x \neq 2k\pi, k \in \mathbb{Z}$. 

My work and thoughts:
If the series is Cesaro summable to $\frac{1}{2}\cot\frac{x}{2}$ for $x \neq 2k\pi, k \in \mathbb{Z}$, this means that if $S_N$ is the $N$th partial sum
then $\sigma_N \rightarrow \frac{1}{2}\cot\frac{x}{2}$, where
$$\sigma_N = \frac{S_1 + \ldots + S_N}{N}.$$
Using trigonometric identities, Euler's formula (and patience) it is not too difficult to show that for the sequence of partial sum we have 
$$S_N = \frac{\cos\frac{x}{2} - \cos(N + \frac{1}{2})x}{2\sin{\frac{x}{2}}}.$$

How do I continue from here to find the given Cesàro sum of the series? 
 A: Using your result, we have
$$\frac1{N} \sum_{k=1}^NS_k = \frac{\cos\frac{x}{2} }{2\sin{\frac{x}{2}}} - \frac{1}{2N\sin{\frac{x}{2}}}\sum_{k=1}^N\cos\left(k + \frac{1}{2}\right)x \\ = \frac1{2}\cot\frac{x}{2} - \frac{1}{2N\sin{\frac{x}{2}}}\sum_{k=1}^N\cos\left(kx + \frac{x}{2}\right) .$$
Now you just need to show that the limit of the sum on the RHS as $N \to \infty$ is zero.
Note that
$$2\sin{\frac{x}{2}}\sum_{k=1}^N\cos\left(kx + \frac{x}{2}\right) = \sum_{k=1}^N2\sin{\frac{x}{2}}\cos\left(kx + \frac{x}{2}\right) \\ = \sum_{k=1}^N \left[\sin(kx+x) - \sin(kx) \right] \\= \sin((N+1)x) - \sin(x),$$
and
$$\lim_{N \to \infty}  \frac{1}{2N\sin{\frac{x}{2}}}\sum_{k=1}^N\cos\left(kx + \frac{x}{2}\right) = \lim_{N \to \infty} \frac{\sin((N+1)x) - \sin(x)}{4N\sin^2{\frac{x}{2}}} \\ = 0.$$
A: There is a more elegant solution that uses divergent series and Euler's Formula $$e^{ix}=\cos x+i\sin x$$
Take the series $$\sum_{k=0}^{\infty} e^{ikx}=\sum_{k=0}^{\infty} (e^{ix})^{k}$$
This is just a geometric series with $x=e^{ix}$. We should note that the series formally is divergent because the geometric series converges only for $|x|<1$ and $|e^{iθ}|=1$. However if one tries to assign a meaningful value to this series we get that
$$\sum_{k=0}^{\infty} (e^{ix})^{k} = \frac{1}{1-e^{ix}}=\frac{1}{1-\cos x-i\sin x}=\frac{1-\cos x+i\sin x}{(1-\cos x)^{2}+\sin^{2}x}=\frac{1-\cos x+i\sin x}{2-2\cos x}$$
$$ \bbox[5px,border:2px solid black] 
{ 
\sum_{k=0}^{\infty} (e^{ix})^{k} =\frac12+i\frac{1}{2}\cot(x/2) 
}$$
So your answer  is easily found by taking the imaginary part of the above expression
Specifically, we found the following interesting relations:
$$\sum_{k=0}^{\infty} \sin(kx) = \frac12\cot(\frac{x}2)$$
and
$$\sum_{k=0}^{\infty} \cos(kx) = \frac12$$
One very interesting question is to wonder why this divergent method and Cesaro summation yield the same result!!
