Edge case with sampling and reconstruction. I know I had been dabbling around this question before, here and here, but does anyone have in their bag of tricks the most simpliest and concise proof that:
$$\sum_{n=-\infty}^{\infty} (-1)^n \, \operatorname{sinc}(t-n) =  \cos(\pi t) $$
where
$$ \operatorname{sinc}(x) \triangleq \begin{cases}
\frac{\sin(\pi x)}{\pi x} \qquad & x \ne 0 \\
\\
1 & x = 0 \\
\end{cases} $$
and $t\in\mathbb{R}$ and $n\in\mathbb{Z}$ ?
I can show that both sides are an even function in $t$ and that both sides have agreement when $t$ is an integer.  But what is the simplest way to show equality for all real $t$ ?
This is something that I want to put together for us Neanderthal electrical engineers.  (and thank you.)
 A: This answer is largely based on this (very concise) answer to a related question of the OP.
Note that for $t\in\mathbb{Z}$ the equality is straightforward to show. The interesting case is when $t$ is not an integer. The derivation below is valid for non-integer real values of $t$.
Using $\cos(x)\sin(y)=\frac12\big[\sin(x+y)-\sin(x-y)\big]$ we can write
$$\begin{align}\sum_{n=-\infty}^{\infty}(-1)^n\textrm{sinc}(t-n)&=\sum_{n=-\infty}^{\infty}\cos(n\pi)\frac{\sin[\pi(t-n)]}{\pi(t-n)}\\&=\frac{\sin(\pi t)}{\pi}\sum_{n=-\infty}^{\infty}\frac{1}{t-n}\\&=\frac{\sin(\pi t)}{\pi}\left[\frac{1}{t}+\sum_{n=1}^{\infty}\left(\frac{1}{t-n}+\frac{1}{t+n}\right)\right]\\&=\frac{\sin(\pi t)}{\pi}\left[\frac{1}{t}+\sum_{n=1}^{\infty}\frac{2t}{t^2-n^2}\right]\tag{1}\end{align}$$
Now we need the following result:
$$\frac{1}{t}+\sum_{n=1}^{\infty}\frac{2t}{t^2-n^2}=\pi\cot(\pi t)\tag{2}$$
which can be found here, here and here, and which can be derived from the well-known infinite product representation of the sinc function
$$\frac{\sin(\pi t)}{\pi t}=\prod_{n=1}^{\infty}\left(1-\frac{t^2}{n^2}\right)\tag{3}$$
Combining $(1)$ and $(2)$ yields the desired result.
A: You should be somewhat careful with how you understand the sum but, assuming that you understand $\sum_{n=-\infty}^\infty a_n$ it as the limit as $N\to\infty$ of $\sum_{-N\le n\le N}(1-\frac{|n|}{N})a_n$ (Cesaro summation, which gives the same result as the usual one when the latter makes sense), you can just write
$$
(-1)^n\rm{sinc}(t-n)=\int_{-1/2}^{1/2}e^{-2\pi i n(x+\frac 12)}e^{2\pi i xt}\,dx
$$
so the Cesaro partial sums become $\int_{-1/2}^{1/2}K_N(x+\frac 12)e^{2\pi i xt}\,dx$ where $K_N(z)=\sum_{-N}^N(1-\frac{|n|}{N}) e^{-2\pi i nz}$ is the Fejer kernel. What you want to know now is that $K_N$ is symmetric, non-negative, $1$-periodic, has total integral $1$ over the period and uniformly tends to $0$ outside an arbitrarily small neighborhood of the integers. So, for large $N$, $K_N(x+\frac 12)$ is a function that is nearly $0$ on $(-\frac 12+\delta,\frac 12-\delta)$ for any fixed $\delta>0$ and has integral nearly $\frac 12$ over each of the intervals $[-\frac 12,-\frac 12+\delta]$ and $[\frac 12-\delta,\frac 12]$. When you integrate anything like that against $e^{2\pi i xt}$ over $[-\frac 12,\frac 12]$, you'll get approximately $\frac 12(e^{-\pi it}+e^{\pi i t})=\cos(\pi t)$.
The only non-pedestrian step in this argument is switching from the usual summation to the Cesaro one. You can avoid it but then you'll get the Dirichlet kernel instead and the last passage to the limit will be somewhat less obvious (the kernel will not decay uniformly in the bulk of the interval but instead it will oscillate faster and faster there and you'll end up using something like the Riemann-Lebesgue lemma to show that you need to look only at the (small neighborhoods of) endpoints.
