This has to do with the Nyquist-Shannon sampling and reconstruction theorem and the so-called Whittaker–Shannon interpolation formula. I had previously asked an ancillary question about this here but this is about a specific nagging issue that seems to "periodically" crop up.
Let's begin with a periodic infinite sequence of real numbers, $a_n \in\mathbb{R}$, having period $N>0\in\mathbb{Z}$. That is:
$$ a_{n+N}=a_n \qquad \forall \ n\in\mathbb{Z}. $$
So there are only $N$ unique values of $a_n$.
Imagine these discrete (but ordered) values as equally spaced on the real number line and being interpolated (between integer $n$) as
$$f(x) = \sum_{n=-\infty}^{\infty} a_n \, \operatorname{sinc}(x-n),$$
where
$$ \operatorname{sinc}(u) \triangleq \begin{cases} \dfrac{\sin(\pi u)}{\pi u} & \text{if } u \ne 0, \\\;1 & \text{if } u = 0. \end{cases} $$
Clearly $f(x)$ is periodic with the same period $N$:
$$ f(x+N) = f(x) \qquad \forall \ x \in \mathbb{R}. $$
All terms are bandlimited to a maximum frequency of $\frac{1}{2}$, so the summation is bandlimited to the same bandlimit. And, in any case, we have
$$ f(x) \Big|_{x = n} = a_n, $$
so the reconstruction works out exactly at the sampling instances.
$$\begin{align} f(x) &= \sum_{n=-\infty}^{\infty} a_n \, \operatorname{sinc}(x-n) \\ &= \sum_{m=-\infty}^{\infty} \sum_{n=0}^{N-1} a_{(n+mN)} \, \operatorname{sinc}\big(x - (n+mN)\big) \\ &= \sum_{m=-\infty}^{\infty} \sum_{n=0}^{N-1} a_n \, \operatorname{sinc}\big(x - (n+mN)\big) \\ &= \sum_{n=0}^{N-1} \left(a_n \, \sum_{m=-\infty}^{\infty} \operatorname{sinc}\big(x - (n+mN)\big)\right). \\ \end{align}$$
Substituting $u \triangleq x-n$ gives
$$ f(x) = \sum_{n=0}^{N-1} a_n \, g(x-n), $$
where
$$ g(u) = \sum_{m=-\infty}^{\infty} \operatorname{sinc}(u-mN). $$
Clearly the continuous (and real) $g(u)$ is periodic with period $N$:
$$ g(u+N) = g(u) \qquad \forall u \in \mathbb{R}. $$
What is the closed-form expression for $g(u)$ in terms of $u$ and $N$?
For $N$ odd, we get the Dirichlet kernel:
$$ g(u) = \frac{\sin(\pi u)}{N \sin(\pi u/N)}. $$
I can then get that expression by an extension of the Discrete Fourier Transform (DFT) and relating it to the continuous Fourier series:
$$ \hat{a}_k \triangleq \frac{1}{\sqrt{N}} \sum\limits_{n=0}^{N-1} a_n \, e^{-i 2 \pi nk/N}, $$
$$ a_n = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} \hat{a}_k \, e^{+i 2 \pi nk/N}. $$
We know that both infinite sequences $a_n$ and $\hat{a}_k$ are periodic with period $N$.
Now, the continuous Fourier series for $f(x)$ is
$$ f(x) = \sum\limits_{k=-\infty}^{\infty} c_k \, e^{+i 2 \pi (k/N) x}, $$
and, because $f(x) \in \mathbb{R}$, we know we have conjugate symmetry
$$ c_{-k} = (c_k)^* \qquad \forall \ k \in \mathbb{Z}. $$
Being "bandlimited" means that
$$ c_k = 0 \qquad \forall \ |k| > \tfrac{N}{2}. $$
But when $N$ is even, what should $g(u)$ be? Now there is potentially a non-zero component to the DFT value at what we EEs call the "Nyquist frequency"; namely $\hat{a}_{N/2}$ exists and might not be zero.
The expression for $g(u)$ I get when $N$ is even is
$$ g(u) = \frac{\sin(\pi u)}{N \tan(\pi u/N)}. $$
But the question is: can it be, in the case that $N$ is even, that
$$ f(x) = \sum_{n=0}^{N-1} a_n \, g(x-n) + A \sin(\pi x),$$
where $A$ can be any real and finite number?
Do you math whiz-bangs know of a good way that I can say, for certain, that $A=0$?
So my most concise question is: for $N$ even and $a_n \in\mathbb{R}$ having period $N>0\in\mathbb{Z}$, namely
$$ a_{n+N}=a_n \qquad \forall \ n\in\mathbb{Z}, $$
is it true that
$$\sum_{n=-\infty}^{\infty} a_n \, \frac{\sin\big(\pi(x-n) \big)}{\pi(x-n)} = \sum_{n=0}^{N-1} a_n \frac{\sin\big(\pi (x-n)\big)}{N \tan\big(\pi (x-n)/N\big)} $$
??
Another way of looking at the question is this special case. Can anyone prove that
$$\sum_{n=-\infty}^{\infty} (-1)^n \, \frac{\sin\big(\pi(x-n) \big)}{\pi(x-n)} = \cos(\pi x) $$
??
The final equation can be written as$$\sum_{n=1}^\infty\frac1{n^2-x^2}=\frac1{2x^2}-\frac\pi{2x}\cot\pi x\quad(x\in\Bbb R\setminus\Bbb Z),$$a proof of which can be found here.