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}


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

  • $\begingroup$ @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. $\endgroup$ – exchange Jun 1 '17 at 7:37
  • $\begingroup$ @exchange , right. That was a silly mistake. Thanks for that. $\endgroup$ – Zaid Alyafeai Jun 1 '17 at 8:27

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.