Sum of infinite series of Sech: $\sum_n \text{sech}(x(n-1/2))\text{sech}(x(n+p-1/2))$ I am wondering if anyone recognises the sum
$\sum_{n=-\infty}^\infty \frac{1}{\cosh\left(x \; (n+p-\frac{1}{2}) \right) \cosh \left(x\;  (n-\frac{1}{2})  \right) }$  ?
I am trying to evaluate sums of this (and related) forms, but I appear to be stuck... 
(The full form I eventually want to find is 
$
  \sum_{n  ,  \vec{p}  \; \in \;\mathbb{Z} \;} 
\;  
 \frac{ 
 \prod_{i}   
 p_i  ^{i-1} 
 \; 
\prod_{i< j }  \; \;  [p_j - p_i]  \; \;
 }{
  \prod_{i}  
  2 \cosh \left[   x \; ( p_i  - \frac{1}{2} )\right]
  \;
   2\cosh \left[  x \; (  p_i+ n  - \frac{1}{2} )\right] 
 }\;
$, but any help would be highly appreciated! )
I have been looking through the papers by Zucker -78 (http://epubs.siam.org/doi/abs/10.1137/0510019) and related works, (such as for example this paper by Bruckman, http://www.fq.math.ca/Scanned/15-4/bruckman.pdf), and know how to evaluate sums of the form $\sum_{n=-\infty}^\infty \frac{1}{ \cosh^2 \left(x\;  (n-\frac{1}{2})  \right) }$, but I run into problems as soon as one of the arguments is shifted.
The answer to the sum above is given in terms of complete elliptic integrals, and I suspect that that will be the case for the shifted one as well, assuming there is a  closed form.
And, on another note, this is my first post here so hope I haven't made any massive mistakes :)
Thanks a lot! 
 A: It is essentially a discrete convolution between $\frac{1}{\cosh}$ and $\frac{1}{\cosh}$. It is interesting to point out that $\frac{1}{\cosh}$ is more or less a fixed point of the Fourier transform:
$$ \mathscr{F}\left(\frac{1}{\cosh x}\right)=\frac{\pi}{\cosh\frac{\pi s}{2}},$$
and its inverse Laplace transform is given by
$$ \mathscr{L}^{-1}\left(\frac{1}{\cosh x}\right) = 2\sum_{n\geq 0}(-1)^n \delta(s-2n-1)=f(s),$$
so:
$$ \frac{1}{\cosh\left(x\left(\frac{1}{2}-n\right)\right)}=\int_{0}^{+\infty}f(s) e^{-sx(n-1/2)}\,ds$$
and:
$$ \frac{1}{\cosh\left(x\left(n-\frac{1}{2}\right)\right)\cosh\left(x\left(n+p-\frac{1}{2}\right)\right)}=\int_{0}^{+\infty}\int_{0}^{+\infty} f(s)\,f(t)\,e^{-ptx}\,e^{-(s+t)x(n+1/2)}\,ds\,dt$$
so:
$$ \sum_{n\geq 0}\frac{1}{\cosh\left(x\left(n-\frac{1}{2}\right)\right)\cosh\left(x\left(n+p-\frac{1}{2}\right)\right)}=\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{ 2\, f(s)\,f(t)\,e^{-ptx}}{\sinh\left((s+t)\frac{x}{2}\right)}\,ds\,dt$$
equals:

$$ \sum_{a\geq 0}\sum_{b\geq 0}\frac{8(-1)^{a+b} e^{-px(2b+1)}}{\sinh((a+b+1)x)}=e^{-px}\sum_{h=0}^{+\infty}\frac{8(-1)^h}{\sinh((h+1)x)}\cdot\frac{1-e^{-2(h+1)px}}{1-e^{-2px}} \tag{1}$$

that is a pretty fast convergent series.

Addendum. We may notice that the degenerate case $p=0$ is way easier to deal with.
Given the Weierstrass product for the $\cosh$ function, we have:
$$ \frac{1}{\cosh^2(x)}=\frac{d^2}{dx^2}\log\cosh(x) = -\sum_{m\geq 0}\left(\frac{1}{\left(x+\pi i\frac{2m+1}{2}\right)^2}+\frac{1}{\left(x-\pi i\frac{2m+1}{2}\right)^2}\right)$$
so we may recognize in the RHS a meromorphic function with equally-spaced and equally-behaving double poles along the imaginary axis. It follows that the meromorphic function defined through
$$ f(x)=\sum_{n\in\mathbb{Z}}\frac{1}{\cosh^2(x+n)} $$
is a constant multiple of a Weierstrass $\wp$ function, whose connection with elliptic integrals is well-known. It is interesting to point out that the RHS of $(1)$, as $p\to 0$, gives a fast-converging expansion for a Weierstrass $\wp$ function. On the other hand, the meromorphic function
$$ g(x)=\sum_{n\in\mathbb{Z}}\frac{1}{\cosh(x+n)\cosh(x+n+t)} $$
has only simple poles, due to:
$$\frac{1}{\cosh x}=\sum_{n\geq 0}(-1)^n\left(\frac{i}{x+(2m+1)\frac{\pi i}{2}}-\frac{i}{x-(2m+1)\frac{\pi i}{2}}\right)$$
so $g(x)$ is related with a Jacobi elliptic function; a useful identity (already exploited in a slightly disguised form through the inverse Laplace transform) is:

$$ \frac{1}{\cosh x} = \frac{2}{\pi}\int_{0}^{+\infty}\frac{\cos\left(\frac{2x}{\pi}u\right)}{\cosh u}\,du\tag{2} $$

that makes us aware of the following fact: a discrete convolution of $\frac{1}{\cosh}$ is the integral over $\mathbb{R}^+$ of $\frac{1}{\cosh u}$ times a discrete convolution of $\cos$, way easier to deal with. It turns out that the RHS of $(1)$ is related with $\frac{1}{\cosh(px)}$, too.
