Pointwise limit of $f_N(x)=\frac{1}{N(x^2+1)}\cdot\sum_{n=1}^N\cos(nx)$

Question

I want to show that the following pointwise limit holds: $$\frac{\sum_{n=1}^N\cos(nx)}{N(x^2+1)}\to\begin{cases} 1, & x=0 \\ 0 & x\neq0 \end{cases}$$ The case $x=0$ is easy: we have, for any $\varepsilon>0$: $$\frac{1}{N}\left\vert\sum_{n=1}^N\cos(nx)-N\right\vert=\left\vert\frac{N}{N}-1\right\vert=0<\varepsilon$$ for any $N$. However, I'm more troubled by the case where $x\neq0$. One issue is that we might have that $x$ is a multiple of $2\pi$, in which case we would have: $$\frac{1}{N(x^2+1)}\left\vert\sum_{n=1}^N\cos(nx)\right\vert=\frac{N}{N(x^2+1)}=\frac{1}{x^2+1}$$ for any $N$, in which case we can't make the above arbitrarily small for $x$ fixed.

Graphically, the pointwise limit seems to hold, but I could be mistaken. Any thoughts?

Answer 1

It doesn't converge to zero for $x=2\pi$ for example:

$\frac{1}{N((2\pi)^2+1)}\cdot \sum\limits_{n=1}^N\cos(n\cdot 2\pi)= \frac{1}{N((2\pi)^2+1)}\cdot \sum\limits_{n=1}^N1=\frac{1}{(2\pi)^2+1}$

Answer 2

Half of the Dirichlet Kernel $$\newcommand{\Re}{\operatorname{Re}} \begin{align} \sum_{n=1}^N\cos(nx) &=\Re\left(\sum_{n=1}^Ne^{inx}\right)\tag1\\ &=\Re\left(\frac{e^{ix}-e^{i(N+1)x}}{1-e^{ix}}\right)\tag2\\ &=\frac{\Re\left(e^{ix}-1-e^{i(N+1)x}+e^{iNx}\right)}{2-2\cos(x)}\tag3\\ &=-\frac12+\frac{\cos(Nx)-\cos((N+1)x)}{2-2\cos(x)}\tag4\\ &=\frac{\sin\left(\left(N+\frac12\right)x\right)}{2\sin\left(\frac12x\right)}-\frac12\tag5 \end{align} $$ Explanation:
$(1)$: $\cos(x)=\Re\left(e^{ix}\right)$
$(2)$: sum of a geometric series
$(3)$: multiply by $\frac{1-e^{-ix}}{1-e^{-ix}}$
$(4)$: $\cos(x)=\Re\left(e^{ix}\right)$
$(5)$: $\cos(Nx)-\cos((N+1)x)=2\sin\left(\left(N+\frac12\right)x\right)\sin\left(\frac12x\right)$

Note that $(5)$ also follows from the Dirichlet Kernel: $$ \sum_{n=-N}^N\cos(nx)=\frac{\sin\left(\left(N+\frac12\right)x\right)}{\sin\left(\frac12x\right)}\tag6 $$

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Application to the Limit

If $\sin\left(\frac12x\right)\ne0$, then, since $\left|\sin\left(\left(N+\frac12\right)x\right)\right|\le1$, $$ \begin{align} \lim_{N\to\infty}\frac1{N\left(x^2+1\right)}\sum_{n=1}^N\cos(nx) &=\lim_{N\to\infty}\frac1{N\left(x^2+1\right)}\left(\frac{\sin\left(\left(N+\frac12\right)x\right)}{2\sin\left(\frac12x\right)}-\frac12\right)\\[6pt] &=0\tag6 \end{align} $$ If $\sin\left(\frac12x\right)=0$, then $x=2\pi k$ for some $k\in\mathbb{Z}$; and thus, $\cos(nx)=1$. Therefore, $$ \begin{align} \lim_{N\to\infty}\frac1{N\left(x^2+1\right)}\sum_{n=1}^N\cos(nx) &=\lim_{N\to\infty}\frac1{N\left(x^2+1\right)}\,N\\[6pt] &=\frac1{x^2+1}\tag7 \end{align} $$