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 $(2)$?
 A: Although I will not give you any solution, nevertheless I will try write an attempt.
I shall start with observation - the value $1/\sqrt{2}$ is rather peculiar for $\arccos$ since $\arccos\frac{1}{\sqrt{2}}=\frac{\pi}{4}$ ...
Define a meromorphic function $$f(z):=\frac{\cot\pi z}{\cosh \pi z + a}$$
Integrating this function along rectangular contour one simply gets the relation :
$$\sum_{n=-\infty}^{\infty}\frac{1}{\cosh\pi n+a}=-2\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{\cosh\pi x-a}-\frac{2}{\sqrt{1-a^2}}\sum_{n=-\infty}^{\infty}\frac{\sinh(2\arccos(-a))}{\cosh 4\pi n-\cosh(2\arccos(-a))}$$
For $a=1/\sqrt{2}$ the second sum is according to Mathematica equal :
$$\frac{Q+2 \pi  \coth \left(\frac{3 \pi }{4}\right)-3\pi}{2 \pi\sinh \left(\frac{3 \pi }{2}\right) }$$
where $Q$ is a constant expressible in terms of Q-Gamma function :
$$Q=\psi _{e^{4 \pi }}\left(\frac{3}{8}\right)-\psi _{e^{4 \pi }}\left(-\frac{3}{8}\right)$$
The integral was calculated already by the user @sirfoga, in our case for the first term : $$-2\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{\cosh\pi x-a}=-\frac{4\arctan{\left(\frac{1+a}{\sqrt{1-a^2}}\right)}}{\pi\sqrt{1-a^2}}$$
For the special case of $a=1/\sqrt 2$ is this equal to $-3/\sqrt 2$ so for the overall sum
$$S=\frac{3}{\sqrt2}-2\sqrt2\coth\frac{3\pi}{4}-\frac{\sqrt2}{\pi}Q=0.75618790046404501626204025904167409716634\dots$$
Connection to the OP's conjecture is at this stage barelly visible :/
Addendum I have run the numbers on Mathematica and it seems the conjecture holds at least for 100 000 digits
A: 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
$$
\frac{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n+\frac{1}{\sqrt{2}}}}{\displaystyle\sum_{n=-\infty}^\infty\frac{1}{\cosh\pi n}}=\frac{S_1(1)}{2K_0/\pi}=\frac{(1+k')}{\sqrt{2}}\left(\frac{2}{\sqrt{1+k'}}-1\right)=\sqrt{2+\sqrt{2}}-\frac{1+\sqrt{2}}{2}.
$$
A: 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}
