Cesàro sum of the series $\sin x + \sin 2x + \sin 3x + \ldots = \frac{1}{2}\cot\frac{x}{2}$ for $x \neq 2k\pi, k \in \mathbb{Z}$

Question

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

$$S_N = \frac{\cos\frac{x}{2} - \cos(N + \frac{1}{2})x}{2\sin{\frac{x}{2}}}.$$

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How do I continue from here to find the given Cesàro sum of the series?

Answer 1

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

$$\lim_{N \to \infty} \frac{1}{2N\sin{\frac{x}{2}}}\sum_{k=1}^N\cos\left(kx + \frac{x}{2}\right) = \lim_{N \to \infty} \frac{\sin((N+1)x) - \sin(x)}{4N\sin^2{\frac{x}{2}}} \\ = 0.$$

Answer 2

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!!