Poisson summation formula clarification regarding Fejer kernel

Define $$\mathbf{F}_R(t) = \begin{cases} R \left(\dfrac{\sin(\pi R t)}{\pi R t}\right)^2 & t \neq 0\\[10pt] R & t = 0 \end{cases} $$

A problem in Stein's Fourier Analysis asks us to prove that the periodization of $\mathbf{F}_R(t)$ is equal to the Fejer kernel on the circle

i.e.

$$\sum_{n=-\infty}^{\infty}\mathbf{F}_N(x+n) = \sum_{n=-N}^{N}\left(1-\frac{|n|}{N}\right)e^{2 \pi i n x} = \frac{1}{N} \frac{\sin^2(N \pi x)}{\sin^2(\pi x)} $$

for $N \geq 1$ an integer

This strongly suggests an application of Poisson summation is needed, which would mean that we need to calculate $$\sum_{n=-\infty}^{\infty}\hat{\mathbf{F}}_N(n)e^{2 \pi i n x}$$

correct?

However, as I don't see how we can go from an infinite series to a finite series using Poisson, I assume we have to show that the above series converges to the closed form expression above? I'm still a bit unclear as to how apply Poisson in this case: if this is correct, any hints as to how to tackle the integral $$\int_{-\infty}^{\infty} \mathbf{F}_N(x)e^{-2 \pi i n x}dx$$ would be appreciated.

How about trying the following alternative approach. First, note that for integer $ N $, the numerator of $ F_N(t) $ factors out since: $$ \sin^2(\pi N (t + n)) = \sin^2(\pi N t + \pi N n) = \sin^2(\pi N t) $$ The remaining part is given by the identity: $$ \sum_{n= -\infty}^{\infty} \frac{1}{\pi^2 (t+n)^2} = \frac{1}{\sin^2(\pi t)} $$ I am not entirely sure how to prove this identity...but I hope this will help anyway.

I know this is an old question, but this was the answer I was looking for, apart from using Exercise#15, although the two questions are very similar and here’s how:

From Exercise#2 it is known that given $g(x) = 1-|x|$ for $|x| \leq 1$, $0$ otherwise, that $\hat{g}(\xi) = \left(\frac{\sin\pi\xi}{\pi\xi}\right)^2$. It can also be calculated relatively easily if you don’t want to take this for granted.

Taking $h(x) = 1 - \frac{|x|}{N}$ for $|x|\leq N$, $0$ otherwise, then $\hat{h}(\xi) = \int_{-N}^{N} \left(1-\frac{|x|}{N}\right) e^{-2\pi i x\xi}dx$, and substitution $u = \frac{x}{N}$, $du = \frac{dx}{N}$, yields $N\cdot \int_{-1}^{1} \left(1-|u|\right) e^{-2\pi i u( \xi N)} \:du = N\cdot \hat{g}(\xi N)= N\cdot \left(\frac{\sin\pi\xi N}{\pi\xi N}\right)^2 = \mathcal{F}_{N}(\xi)$ if $\xi \not= 0$, as seen in Exercise#9. By the Fourier Inversion formula, $\hat{\hat{h}}(x) = h(-x) = h(x)$ since $h(x)$ is even. By Poisson (the summation formula),

$\sum_\limits{n=-\infty}^{\infty}\mathcal{F}_N (x+n)= \sum\limits_{n=-\infty}^\infty h(n)e^{2\pi i nx}=\sum\limits_{n=-N}^N \left(1-\frac{|n|}{N}\right)e^{2\pi i nx}$

going from an infinite series to a finite series because $h(x)$ is zero for $|x| > N$.