# 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.

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

$$\prod_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n} e^{2} = \sqrt{2} \mathrm{exp} \left\{ \frac{2K}{\pi}-1 \right\}$$

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

$$4 \int^{1/4}_0 x\pi \cot(\pi x) \, dx -1= 4 \left(\frac{\log(2\sin\pi /4)}{4}+\frac{\mathrm{cl}_2(\pi/2)}{2\pi}\right)-1 = \frac{\log 2}{2}+\frac{K}{2\pi}-1$$

• @Zaid: Sorry to post something unrelated to the question but I don't know how to contact you otherwise: I just downloaded your book you linked on your stackexchange profile and think that there is a slight typo on the first page, namely it should be $F=2^a~\Rightarrow~F'(a)=\log(2)2^a$ instead of $F'(a)=\log(a)2^a$, right? Great work the book. – exchange Jun 1 '17 at 7:37
• @exchange , right. That was a silly mistake. Thanks for that. – Zaid Alyafeai Jun 1 '17 at 8:27