How do we prove this asymptotic relation $\lim\limits_{k\rightarrow \infty}G(k)=\sqrt{2}$ I made the following observation with Mathematica:
Consider the infinite series,for natural number $k$
$$G(k)=\sum_{n=1}^{\infty}\frac{\sinh^{n}(\sqrt{k}\pi)}{\sqrt{(k+1)+\cosh{(2\sqrt{k}\pi n)}}}$$
then
$$\lim\limits_{k\rightarrow \infty}G(k)=\sqrt{2}$$

How do we prove this asmptotic relation $G(k)\sim \sqrt{2}$?

 A: Let $m=\sqrt{k}$. Using the exponential representations of $\sinh$ and $\cosh$, followed by the binomial theorem on the numerator of $a_n$.
$$\begin{aligned}
a_n&=\frac{\sinh^{n}(\sqrt{k}\pi)}{\sqrt{(k+1)+\cosh{(2\sqrt{k}\pi n)}}}\\
&=\frac{\sqrt{2}}{2^n}\frac{\left(e^{\pi m}-e^{-\pi m}\right)^n}{\sqrt{2(m^2+1)+e^{2\pi nm}+e^{-2\pi nm}}}\\
\\
&=\frac{\sqrt{2}}{2^n}\frac{\sum_{i=0}^n (-1)^i{n \choose i}e^{\pi m (n-2i)}}{\sqrt{2(m^2+1)+e^{2\pi nm}+e^{-2\pi nm}}}\\
\\
&=\frac{\sqrt{2}}{2^n}\sum_{i=0}^n(-1)^i{n \choose i}\underbrace{\left(\frac{ e^{\pi m (n-2i)}}{\sqrt{2(m^2+1)+e^{2\pi nm}+e^{-2\pi nm}}}\right)}_{b_i}
\\
\lim_{m\to\infty}a_n&=\lim_{m\to\infty}\frac{\sqrt{2}}{2^n}\sum_{i=0}^n(-1)^i{n \choose i}b_i\\
&=\frac{\sqrt{2}}{2^n}\quad{\text{by $(*)$}}\\
\end{aligned}$$
Therefore, as we can exchange the limits $k\to\infty$ and $m\to\infty$,
$$\begin{aligned}
\lim_{k\to\infty}G(k)&=\sum_{n=1}^{\infty}\lim_{k\to\infty}{a_n}\\
&=\sum_{n=1}^{\infty}\frac{\sqrt{2}}{2^n}\\
&=\sqrt{2}
\end{aligned}$$


Theorem: 
  $$\lim_{m\to \infty} b_i=\begin{cases}1&i=0\\0&\text{else}\end{cases}\tag{*}$$

Proof
At $i=0$,
$${b_i}^2=\frac{1}{1+\frac{\left(2\left(m^2+1\right)+e^{-2\pi mn}\right)}{e^{2\pi mn}}}$$
So, as $m\to\infty$, $e^{2\pi mn}$ dominates the other terms and ${b_i}^2\to\frac{1}{1+0}=1$.
Conversely, since $2(m^2+1)+e^{-2\pi nm}>0$, $${b_i}^2=\frac{e^{2\pi m\left(n-2i\right)}}{e^{2\pi nm}+[2(m^2+1)+e^{-2\pi nm}]}\leq\frac{e^{2\pi m\left(n-2i\right)}}{e^{2\pi nm}}=\frac{1}{e^{4\pi im}}$$
So for all $i>0$, $$0\leq \lim_{m\to \infty} {b_i}^2\leq \lim_{m\to\infty}\frac{1}{e^{4\pi im}}=0$$
Then, $\lim_{m\to \infty} b_i=0$ by the squeeze theorem.  
