Conjectured closed form for $\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}$ 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 $(2)$?

Disclaimer: I verified my self that $\frac{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}} \approx 0.640652 \approx \sqrt{\sqrt{2}+2}-\frac{1}{2} \left(\sqrt{2}+1\right)$ so mine is only an approximation but as far as I could do, no closed form exists to evaluate precisely your sum: besides, if your aim is to do computational evaluations, I suggest to simplify the sum, evaluate the integral I'm providing or evaluate the sum from $n = -10^3$ to $n = 10^3$: I've seen no massive difference with $n = -10^6$ to $n = 10^6$ or more.
\begin{align}
\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}} &\approx \int_{-\infty}^\infty\frac{1}{\cosh\pi x+\frac{1}{\sqrt{2}}} dx\\
&= -\frac{2 \sqrt{2} \arctan\left(\frac{(\sqrt{2}-2) \tanh \left(\frac{\pi  n}{2}\right)}{\sqrt{2}}\right)}{\pi } \big|_{-\infty}^{+\infty}\\
&= \sqrt{2}
\end{align}
and more in general
\begin{align}
\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+a} &\approx \int_{-\infty}^\infty\frac{1}{\cosh\pi x+a} dx\\
&= -\frac{2 \arctan\left(\frac{(a-1) \tanh \left(\frac{\pi  n}{2}\right)}{\sqrt{1-a^2}}\right)}{\pi  \sqrt{1-a^2}} \big|_{-\infty}^{+\infty}\\
&= \left( -\frac{2 \arctan\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\pi  \sqrt{1-a^2}} \right) - \left(\frac{2 \arctan\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\pi  \sqrt{1-a^2}} \right)\\
&= -2 \left(\frac{2 \arctan\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\pi  \sqrt{1-a^2}} \right)
\end{align}
so in your specific case
\begin{align}
\frac{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}} &\approx \frac{\int_{-\infty}^\infty\frac{1}{\cosh\pi x+\frac{1}{\sqrt{2}}} dx}{\int_{-\infty}^\infty\frac{1}{\cosh\pi x} dx}\\
&= \frac{-2 \left(\frac{2 \arctan\left(\frac{\frac{1}{\sqrt{2}}-1}{\sqrt{1-\frac{1}{\sqrt{2}^2}}}\right)}{\pi  \sqrt{1-\frac{1}{\sqrt{2}^2}}} \right)}{-2 \left(\frac{2 \arctan\left(\frac{0-1}{\sqrt{1-0^2}}\right)}{\pi  \sqrt{1-0^2}} \right)}\\
&= \frac{1}{\sqrt{2}}\\
&\approx 0.707107
\end{align}
