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)

• $\sum_{n=1}^{\infty} \text{sech} \left( \pi n \right)$ might be more interesting since it has a nice closed form solution in terms of the gamma function. Wolfram Alpha express your sum in terms of some strange form of the digamma function. Feb 9 '13 at 15:43
• @Random Variable: interesting. Do you know how to compute your version? Feb 9 '13 at 15:49

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.

• +1 thanks for sharing the connection to elliptic functions. Feb 9 '13 at 20:55

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.

• (+1) interesting and nice. Looking at your work and at what RandomVariable said, I wonder if the problem I received had an error and there was a $\pi$ instead of $2$. Feb 9 '13 at 19:50