Compute $ \sum_{k=1}^{\infty} \text{sech}(2 k)$ How would would you approach the below series?
$$ \sum_{k=1}^{\infty}  \text{sech}(2 k)$$
Thanks in advance for your hints, suggestions. (I need a starting point - Sister) 
 A: This can be expressed in terms of a Lambert series, which will then make it clear why q-digamma functions or Jacobi theta functions are the right functions to consider. Write $\mbox{sech}(2k)=\frac{2}{\exp(2k)+\exp(-2k)}=\frac{2\exp(-2k)}{1+\exp(-4k)}$. Now let $\beta=\exp{-2}$ and notice that $|\beta|<1$. So your series can be written as:
$$\sum_{k=1}^\infty \mbox{sech}(2k)=2\sum_{k=1}^\infty \frac{\beta^k}{1+\beta^{2k}}=\frac{1}{2}\left[\vartheta_3^2(\beta)-1\right]$$
where $\vartheta_3(\beta)=\sum_{q=-\infty}^\infty \beta^{n^2}$ is the Jacobi theta function (whose square happens to count the number of ways to write a number as the sum of two squares). To see the above identity, write the above sum as : $\sum_{k=1}^\infty \beta^k\sum_{n=0}^\infty (-1)^n\beta^{2nk}$ and reverse, taking care to group the appropriate terms.
A: Consider a function $f(a) = \sum_{k=1}^\infty \frac{1}{\cosh(a k)}$, so 
that the quantity of interest equals $f(2)$. Rewriting hyperbolic 
function using exponentials we have
$$
    f(a) = \sum_{k=1}^\infty \frac{2 \exp(-a k)}{1+\exp(-2 a k)}
$$
Denoting $q=\exp(-a)$ we write:
$$
    f(a) = \sum_{k=1}^\infty \frac{2 q^k}{1+q^{2k}} = \sum_{k=1}^\infty 
\frac{(-1,q^2)_{k}}{(-q^2,q^2)_{k}} q^{k} = {}_2 
\phi_1\left(\left.\begin{array}{cc}q^2 &-1 \cr -q^2 & \end{array} 
\right| q^2; q\right) - 1
$$
where ${}_2\phi_1\left(\left.\begin{array}{cc}a_1 &a_2 \cr b_1 & 
\end{array} \right| q; z\right)$ denotes a basic hypergeometric 
series.
Here is a numerical confirmation in Mathematica:
 In[197]:= fapprox[a_] := Sum[Sech[a k], {k, 1, 1000}];

 In[201]:=
 fexact[a_] :=
   QHypergeometricPFQ[{Exp[-2 a], -1}, {-Exp[-2 a]}, Exp[-2 a],
     Exp[-a]] - 1;

 In[202]:= With[{a = Pi/4}, N[fexact[a], 50] == fapprox[a]]

 Out[202]= True

Per Mathematica this sum has a closed-form result for $a=\pi$, as 
alluded to in the comments by @RandomVariable:
 In[203]:= Sum[Sech[Pi*n], {n, Infinity}]

 Out[203]= (1/2)*(-1 + Sqrt[Pi]/Gamma[3/4]^2)

and likely getting this result requires a detour into elliptic functions.
