prove that: $\sqrt{2}=e^{1-{2K\over \pi}}\prod\limits_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n}e^2$

Question

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.

Answer 1

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

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

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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

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

Answer 2

Consider the $N$-th partial summation,

\begin{align*}S_N&=\sum\limits_{n=1}^{N} \left(2n\log \left(\frac{4n+1}{4n-1}\right) - 1\right) \\&= -N -\frac{1}{2}\sum\limits_{n=1}^{N} \log [(4n+1)(4n-1)] + \frac{1}{2}\sum\limits_{n=1}^{N} (4n+1)\log (4n+1) - (4n-1)\log (4n-1)\\&= -N - \frac{1}{2}\log \frac{(4N+1)!}{2^{2N}(2N)!}+\frac{1}{2}\sum\limits_{n=1}^{2N}(-1)^n (2n+1)\log (2n+1)\\&= -N + N\log 2 - \frac{1}{2}\log \frac{(4N+1)!}{2^{2N}(2N)!}+\sum\limits_{n=1}^{2N}(-1)^{n} \left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right)\\&\underbrace{\sim}_{\substack{N \to \infty\\\text{Stirling Approx.}}} - \frac{1}{2}\log \frac{2^{1/2}(4N+1)^{2N+1}}{e^{1/2}2^{2N}}+\sum\limits_{n=1}^{2N}(-1)^{n} \left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right)\\&= \frac{1}{4}-\frac{1}{4}\log 2 +\sum\limits_{n=0}^{2N}(-1)^{n} \left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right)-\left(N+\frac{1}{2}\right)\log \left(2N+\frac{1}{2}\right)\end{align*} Also note the telescopic summation: $\displaystyle \frac{1}{2}\sum\limits_{n=0}^{2N} (-1)^n\left(n\log n + (n+1)\log (n+1)\right) = \left(N+\frac{1}{2}\right)\log (2N+1)$

Hence, \begin{align*}S_N &= \frac{1}{2} - \frac{1}{4}\log 2 + \sum\limits_{n=0}^{2N} (-1)^n\left(\left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right) - \frac{1}{2}n\log n - \frac{1}{2}(n+1)\log (n+1)\right) + o(1)\end{align*} Now, $\displaystyle \lim\limits_{s \to 0} \frac{1}{s}\left(\frac{1}{\left(n+\frac{1}{2}\right)^{s-1}} - \frac{1}{2n^{s-1}}-\frac{1}{2(n+1)^{s-1}}\right) = -\left(\left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right) - \frac{1}{2}n\log n - \frac{1}{2}(n+1)\log (n+1)\right)$

as a result we have, \begin{align*}&\sum\limits_{n=0}^{2N} (-1)^n\left(\left(n+\frac{1}{2}\right)\log \left(n+\frac{1}{2}\right) - \frac{1}{2}n\log n - \frac{1}{2}(n+1)\log (n+1)\right)\\&= \lim\limits_{s \to 0} \frac{1}{s}\sum\limits_{n=0}^{2N} (-1)^n\left(\frac{1}{2n^{s-1}}+\frac{1}{2(n+1)^{s-1}}-\frac{1}{\left(n+\frac{1}{2}\right)^{s-1}}\right)\\&= \lim\limits_{s \to 0} \frac{1}{s}\sum\limits_{n=0}^{2N} \frac{(-1)^n}{\Gamma(s-1)}\int_0^{\infty} \left(\frac{1}{2}+\frac{e^{-x}}{2} - e^{-x/2}\right)x^{s-2}e^{-nx}\,dx \\&= -\sum\limits_{n=0}^{2N} (-1)^n\int_0^{\infty} \left(\frac{1}{2}+\frac{e^{-x}}{2} - e^{-x/2}\right)x^{-2}e^{-nx}\,dx\end{align*} Hence, $\displaystyle \lim\limits_{N \to \infty} S_N = \frac{1}{2}-\frac{1}{4}\log 2 - \frac{1}{2}\int_0^{\infty} \frac{1+e^{-x}-2e^{-x/2}}{x^2(1+e^{-x})}\,dx$

Now, to evaluate the above integral we use an indirect approach.

The following identity is well known (easy to prove): $\displaystyle \int_0^{\infty} \frac{t^{-s}s}{1+t}\,dt = \frac{\pi s}{\sin \pi s}$ and $\displaystyle \int_0^{1/2} \frac{\pi s}{\sin \pi s}\,ds = \frac{2G}{\pi}$

As a result we might write: \begin{align}\frac{2G}{\pi} &= \int_0^{\infty} \int_0^{1/2} \frac{t^{-s}s}{1+t}\,ds \,dt \\&= \int_0^{\infty} \frac{1- t^{-1/2} - \frac{1}{2}t^{-1/2}\log t}{(t+1)\log^2 t}\,dt\\&= \int_{-\infty}^{\infty} \frac{e^{-x}- e^{-x/2} + \frac{1}{2}xe^{-x/2}}{(1+e^{-x})x^2}\,dx \\&= \int_{0}^{\infty} \frac{e^{-x}- e^{-x/2} + \frac{1}{2}xe^{-x/2}}{(1+e^{-x})x^2}\,dx + \int_{-\infty}^{0} \frac{e^{-x}- e^{-x/2} + \frac{1}{2}xe^{-x/2}}{(1+e^{-x})x^2}\,dx \\&= \int_0^{\infty} \frac{1+e^{-x}-2e^{-x/2}}{x^2(1+e^{-x})}\,dx\end{align} where, we made the substitution $t = e^{-x}$ in step $(2)$.

Combining the results, $$\sum\limits_{n=1}^{\infty} \left(2n\log \left(\frac{4n+1}{4n-1}\right) - 1\right) = \frac{1}{2}-\frac{1}{4}\log 2 -\frac{G}{\pi}$$