Show that $\frac{1}{2} + \sum_{j=1}^\infty \cos(jx) = 0$ in the Cesàro sense

Question

So, we have $s_n = \frac{1}{2} + \sum_{j=1}^n \cos(jx)$ and we wish to show that $s_n = 0$ in the Cesàro sense. My understanding of the Cesàro method is to take $\frac{\sum_{n=1}^N s_n}{N}$ which yields $\frac{1}{2N} + \frac{\sum_{n=1}^N \cos(nx)}{N} \rightarrow 0$ as $N \rightarrow \infty$. Is this correct?

Answer 1

To show that $\frac{1}{2} + \sum_{j=1}^{\infty}\cos(jx)$ is Cesaro summable to $0$ you must show that the average value of partial sums converges to $0$, that is

$$\lim_{n \to \infty} \frac1{n}\sum_{k=1}^nS_k = 0,$$

where

$$S_k = \frac{1}{2} + \sum_{j=1}^{k}\cos(jx)= \frac{\sin[(k + 1/2)x]}{2 \sin(x/2)} .$$

Hence,

$$\frac1{n} \sum_{k=1}^nS_k = \frac1{n}\sum_{k=1}^n\frac{\sin[(k + 1/2)x]}{2 \sin(x/2)}.$$

It can be shown that

$$\sum_{k=1}^n \sin[(k + 1/2)x] = \frac{\sin(nx/2)\sin[(n+2)x/2]}{\sin(x/2)}.$$

Hence,

$$\begin{align} \frac1{n} \sum_{k=1}^nS_k &= \frac1{2n}\frac{\sin(nx/2)\sin[(n+2)x/2]}{\sin^2(x/2)}\end{align},$$

and the RHS converges to $0$ as $ n \to \infty$ for $x$ not equal to a multiple of $2\pi$.