Show that the sum equals $0$ according to Cesàro

Question

I'm stuck at some problems in my Fourier Analys course, maybe you got a clue.

If $x \neq n \cdot 2 \pi$, then

$$ \frac{1}{2}+ \sum_{k=1}^{\infty}\cos(kx) = 0, \quad (C,1). $$

Solution:

So I know where I want to go and there is only one step I fail with when I peak at the solution manual, here we go

$$ \frac{1}{2}+ \sum_{k=1}^{\infty}\cos(kx) = 0 \Leftrightarrow 1+ 2\sum_{k=1}^{\infty}\cos(kx) = 0 $$

Now rewrite $\cos(kx)= \frac{1}{2}(e^{ikx}+e^{-ikx})$ and we have that

$$ 1+ \sum_{k=1}^{\infty}(e^{ikx}+e^{-ikx}) = 0 $$

and now the solution manual shows that the sum equals $-1$ and we are done, anyone know how to do that?

Answer 1

As shown in Nicolas' answer, the sum $\sum_{k=1}^n\cos(kx)$ is given by

$$\sum_{k=1}^n\cos(kx)=\frac{\cos(nx)-\cos((n+1)x)}{2(1-\cos(x))}-\frac12,\qquad x\neq 2\pi l\tag{1}$$

The Cesaro sum is

$$\lim_{m\rightarrow\infty}\frac{1}{m}\sum_{n=1}^m\left[\frac{\cos(nx)-\cos((n+1)x)}{2(1-\cos(x))}-\frac12\right]=\\=\lim_{m\rightarrow\infty}\frac{1}{m}\frac{\cos(x)-\cos((m+1)x)}{2(1-\cos(x))}-\frac12=-\frac12$$

Answer 2

You are on the good track.

$$1+ \sum_{k=1}^{\infty}(e^{ikx}+e^{-ikx}) = 1+ \sum_{k=1}^{\infty}e^{ikx}+\sum_{k=1}^{\infty}e^{-ikx}$$ Now, $$e^{ikx}=\big(e^{ix}\big)^k$$ $$e^{-ikx}=\big(e^{-ix}\big)^k$$ So, two geometric series, isn't it ?

I am sure that you can take it from here.

Answer 3

EDIT : This post is not a full answer, the conclusion is given in Matt L.'s post.

Let us compute \begin{align}\sum_{k=1}^{n}\mathrm{e}^{ikx}+\sum_{k=1}^{n}\mathrm{e}^{-ikx} & = \sum_{k=0}^{n}\mathrm{e}^{ikx}-1+\sum_{k=0}^{n}\mathrm{e}^{-ikx}-1 \\ & = \frac{1-\mathrm{e}^{i(n+1)x}}{1-\mathrm{e}^{ix}}-1+\frac{1-\mathrm{e}^{-i(n+1)x}}{1-\mathrm{e}^{-ix}}-1 \\ & = \frac{\mathrm{e}^{ix}-\mathrm{e}^{i(n+1)x}}{1-\mathrm{e}^{ix}}+\frac{\mathrm{e}^{-ix}-\mathrm{e}^{-i(n+1)x}}{1-\mathrm{e}^{-ix}} \\ & = \frac{\mathrm{e}^{ix}-\mathrm{e}^{i(n+1)x}-1+\mathrm{e}^{inx}+\mathrm{e}^{-ix}-\mathrm{e}^{-i(n+1)x}-1+\mathrm{e}^{-inx}}{1-\mathrm{e}^{ix}+1-\mathrm{e}^{-ix}} \\ & = \frac{2\left(-1+\cos(x)+\cos(nx)-\cos((n+1)x)\right)}{2\left(1-\cos(x)\right)} \\ & = -1+\frac{\cos(nx)-\cos((n+1)x)}{1-\cos(x)}. \end{align}