TITLE: Conjectured closed form for $\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}$
QUESTION [28 upvotes]: I was trying to find closed form generalizations of the following well known hyperbolic secant sum
$$
\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}=\frac{\left\{\Gamma\left(\frac{1}{4}\right)\right\}^2}{2\pi^{3/2}},\tag{1}
$$
as
$$
S(a)=\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+a}.
$$
In particular I find by numerical experimentation
$$
\displaystyle \frac{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}}\overset{?}=-\frac{1}{2}\left(1+\sqrt{2}\right)+\sqrt{2+\sqrt{2}}\tag{2}
$$
(Mathematica wasn't able to find a closed form directly, but then I decided to switch to calculation of ratios of the sums, calculated ratios numerically and then was able to recognize this particular ratio as a root approximant. This was subsequently verified to 1000 decimal places).
I simplified this expression from the previous edition of the question.
Unfortunately for other values of $a$ I couldn't find a closed form. Of course $(2)$ together with $(1)$ would imply a closed form for the sum $S(1/\sqrt{2})$
How one can prove $(2)$?

REPLY [14 votes]: Let
$$
S_1(\alpha)=\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi \alpha n+\frac{1}{\sqrt{2}}},
$$
$$
S_2(\alpha)=\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi \alpha n-\frac{1}{\sqrt{2}}},
$$
then due to $2\cosh^2x-1=\cosh 2x$ one obtains
$$
S_2(\alpha)-S_1(\alpha)=2\sqrt{2}\sum_{n=-\infty}^\infty\frac{1}{\cosh 2\pi \alpha n},
$$
$$
S_2(\alpha)+S_1(\alpha)=4\sum_{n=-\infty}^\infty\frac{\cosh\pi\alpha n}{\cosh 2\pi \alpha n}.
$$
Now if one defines elliptic integrals of the first kind $K$ and $\Lambda$ according to equations $\frac{K'}{K}=\frac{K(k')}{K(k)}=\alpha$, $\frac{\Lambda'}{\Lambda}=\frac{K(k_1')}{K(k_1)}=2\alpha$, where $k'=\sqrt{1-k^2},~k_1'=\sqrt{1-k_1^2}$, then the well known formulas from the theory of elliptic functions (see Whittaker and Watson, A Course of Modern Analysis) state that
$$
\sum_{n=-\infty}^\infty\frac{1}{\cosh \pi \alpha n}=\frac{2K}{\pi},~\sum_{n=-\infty}^\infty\frac{1}{\cosh 2\pi \alpha n}=\frac{2\Lambda}{\pi},~\sum_{n=-\infty}^\infty\frac{\cosh\pi\alpha n}{\cosh 2\pi \alpha n}=\frac{2\Lambda}{\pi}~\text{dn}(i\Lambda'/2,k_1),
$$
$$
k_1=\frac{1-k'}{1+k'},\quad \Lambda=\frac{1}{2}(1+k')K,\quad \text{dn}(i\Lambda'/2,k_1)=\sqrt{1+k_1}.
$$
From this by trivial algebra one can deduce that

$$
S_1(K'/K)=\frac{K\sqrt{2}}{\pi}(1+k')\left(\frac{2}{\sqrt{1+k'}}-1\right).
$$

Now for $k=1/2$ one has $k'=1/2$, $K=K'=K_0$, therefore
$$
\frac{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}}=\frac{S_1(1)}{2K_0/\pi}=\frac{(1+k')}{\sqrt{2}}\left(\frac{2}{\sqrt{1+k'}}-1\right)=\sqrt{2+\sqrt{2}}-\frac{1+\sqrt{2}}{2}.
$$