prove that: $\sqrt{2}=e^{1-{2K\over \pi}}\prod\limits_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n}e^2$ show that
$$\sqrt{2}=e^{1-{2K\over \pi}}\prod_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n}e^2$$
where K is the catalan's constant; $K=0.9156 ...$
My try:
take the ln
$${1\over2}\ln{2}=\left(1-{2K\over \pi}\right)\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n}}e^2$$
$${1\over2}\ln{2}=\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n}}e^2-
{2K\over \pi}\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n}}e^2$$
we know that
$${1\over 2}\ln{2}=\sum_{n=1}^{\infty}\ln{\left(4n-1\over 4n+1\right)}+\sum_{n=1}^{\infty}\ln{\left(4n+1\over 4n+2\right)}$$
sub: then we got
$$\sum_{n=1}^{\infty}\ln{\left(4n+1\over 4n+2\right)}=\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n-1}}e^2-
{2K\over \pi}\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n}}e^2$$
Anyway I am stuck, any help please. I tried and looked everywhere on wolfram can't find any similiar infinite product to simplify this further. 
 A: Consider 
\begin{align}
\sum_{n=1}^{\infty}4n \log\left({4n-1\over 4n+1}\right)+2 &=-
\sum_{n=1}^{\infty}4n \sum_{k=0}^\infty \frac{-2}{(2k+1)(4n)^{2k+1}}+2\\
&=-\sum_{k=1}^\infty \frac{2}{(2k+1)4^{2k}}\sum_{n=1}^{\infty}\ \frac{1}{n^{2k}}\\ 
&=-\sum_{k=1}^\infty \frac{2\zeta(2k)}{(2k+1)4^{2k}}\\
&= \frac{2K}{\pi}-1+\frac{\log(2)}{2}
\end{align}
Hence finally we have 
$$\prod_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n} e^{2} = \sqrt{2} \mathrm{exp} \left\{ \frac{2K}{\pi}-1 \right\}$$

ADDENDUM 
We prove the last series using the generating function
$$\pi\;x\;\cot(\pi\;x)-1=-2\sum_{k=1}^\infty \zeta(2k)\;x^{2k}$$
By integration 
$$4\pi\int^{1/4}_0\;x\;\cot(\pi\;x)\,dx -1=-2\sum_{k=1}^\infty \zeta(2k)\frac{x^{2k+1}}{(2k+1) 4^{2k}}$$
Note that 
\begin{align}
    \int^z_0 x\pi \cot(\pi x) \, dx 
    &=z\log(\sin\pi z)-\int^z_0 \log(\sin\pi x) dx\\
    &=z\log(2\sin\pi z)-\int^z_0 \log(2\sin\pi x) dx\\
    &=z\log(2\sin\pi z)-\frac{1}{2\pi }\int^{2\pi z}_0 \log\left(2\sin\frac{x}{2}\right) dx\\
    &=z\log(2\sin\pi z)+\frac{\mathrm{cl}_2(2\pi z)}{2\pi}\\
    \end{align}
Hence 
$$ 4  \int^{1/4}_0 x\pi \cot(\pi x) \, dx -1= 4 \left(\frac{\log(2\sin\pi /4)}{4}+\frac{\mathrm{cl}_2(\pi/2)}{2\pi}\right)-1 = \frac{\log 2}{2}+\frac{K}{2\pi}-1$$
