Show that, $\sum_{n=1}^{\infty}\left[\frac{\beta(2n)}{n}-\ln\left(\frac{n+1}{n}\right)\right] =...$ valid for all $s\ge 1$
$$\beta(s)=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}$$
The particular value of $\Gamma\left(\frac{1}{4}\right)=3.6256099...$
Euler's constant is defined by $$\lim_{n \to \infty}\left[H_n-\ln(n)\right]=\gamma$$
Euler also showed that $$\gamma=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\ln\left(\frac{n+1}{n}\right)\right)$$
Where $H_n=\sum_{k=1}^{n}\frac{1}{k}$
Show that,
$$\sum_{n=1}^{\infty}\left[\frac{\beta(2n)}{n}-\ln\left(\frac{n+1}{n}\right)\right] =\gamma+\ln\left(\frac{16\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}\right)$$
We are greatly appreciated if anyone can answer this  identity.
 A: With the same approach to a recent question of yours,
$$ \beta(2n)-1=\frac{1}{\Gamma(2n)}\int_{0}^{+\infty}\frac{x^{2n-1} e^{-3x}}{1+e^{-2x}}\,dx $$
hence:
$$ \sum_{n\geq 1}\frac{\beta(2n)-1}{n} = 2\int_{0}^{+\infty}\frac{e^{-3x}}{1+e^{-2x}}\cdot\frac{\cosh x-1}{x}\,dx.$$
By Frullani's theorem, or differentiation under the integral sign (Feynman's trick):
$$ 2\int_{0}^{+\infty}\frac{\cosh x-1}{x}e^{-(3+2m)x}\,dx =-\log(m+1)+2\log\left(m+\frac{3}{2}\right)-\log(m+2),$$
hence:
$$\sum_{n\geq 1}\frac{\beta(2n)-1}{n} = \sum_{m\geq 0}(-1)^m\left(-\log(m+1)+2\log\left(m+\frac{3}{2}\right)-\log(m+2)\right)$$
is the logarithm of an infinite product that can be computed from the limit product representation for the $\Gamma$ function. Namely:
$$ \sum_{m\geq 0}\left[\log\left(2m+\frac{3}{2}\right)-\frac{1}{2}\log(2m+1)-\log\left(2m+\frac{5}{2}\right)+\frac{1}{2}\log(2m+3)\right]\\=\frac{\log 2}{2}-\log\Gamma\left(\frac{3}{4}\right)+\log\Gamma\left(\frac{5}{4}\right).$$
Now it is enough to use the $\Gamma$ reflection formula and the identity $\Gamma(z+1)=z\,\Gamma(z)$ to express your sum in terms of $\gamma$ and $\Gamma\left(\frac{1}{4}\right)$. The original identity is, in fact, equivalent to:

$$ \prod_{m\geq 0}\frac{(4m+3)(4m+3)(4m+6)}{(4m+2)(4m+5)(4m+5)}=\color{red}{\frac{1}{16\pi^2}\,\Gamma\left(\frac{1}{4}\right)^4}.$$

