$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}?$ 
$$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\left(\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}\right)\tag1$$

Where $\beta(n)$ is Beta dirichlet function
$(1)$ becomes
$$ln\left({2\over \pi}\right)+{1\over n\Gamma(n)}\sum_{n=1}^{\infty}(-1)^{n-1}\int_{0}^{\infty}{x^{n-1}\over e^x+e^{-x}}\mathrm dx\tag2$$
How can we show that the closed form for $(1)$ is correct?
 A: Let us consider the two sums separately:
$$
S_1 = \sum_{n=1}^\infty (-1)^{n-1}\frac{\beta(n)}{n} \\
S_2 = \sum_{n=1}^\infty (-1)^{n-1}\ln\frac{n+1}{n}
$$
For $S_1$, let us first consider the related sum, for which $S_1$ is the limiting value $S(1)$:
$$
S(x) = \sum_{n=1}^\infty (-1)^{n-1}\frac{\beta(n)x^n}{n} \\
S'(x) = \sum_{n=1}^\infty (-1)^{n-1}\beta(n)x^{n-1} = \sum_{n=1}^\infty (-1)^{n-1}x^{n-1}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^n} \\
= \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \sum_{n=1}^\infty \left( \frac{-x}{2k+1} \right)^{n-1} = \sum_{k=0}^\infty \frac{1}{2k+1+x} = \frac{1}{2} \Phi \left(-1, 1, \frac{1+x}{2} \right)
$$
Where $\Phi$ is the Lerch Transcendent, and $|x|<1$. Now, upon integration, we can recover $S_1$:
$$
S_1 - S(0) =\int_0^1 S'(x) dx = \int_0^1 \frac{1}{2} \Phi \left(-1, 1, \frac{1+x}{2} \right)dx = \int_\frac{1}{2}^1 \Phi (-1, 1, x)dx \\
= \lim_{s\to0}\int_\frac{1}{2}^1 \Phi (-1, 1+s, x)dx = \lim_{s\to0} \left[ \frac{-1}{s}\Phi (-1, s, x) \right|_\frac{1}{2}^1 \\
= \lim_{s\to0} \frac{\Phi \left(-1, s, \frac{1}{2}\right) - \Phi(-1, s, 1)}{s} = \lim_{s\to0} \frac{2^s \beta(s) - \eta(s)}{s} \\
=^H \lim_{s\to0} \ \ln(2) 2^s \beta(s) + 2^s \beta'(s) - \eta'(s) = \frac{1}{2} \ln(2) + \ln \frac{\Gamma^2(\frac{1}{4})}{2 \pi \sqrt{2}} - \frac{1}{2} \ln \frac{\pi}{2} = \ln \frac{\sqrt{2 \pi}}{\Gamma^2(\frac{3}{4})}
$$
And as $S(0) = 0$, this is all $S_1$. For $S_2$:
$$
S_2 = \sum_{n=1}^\infty (-1)^{n-1}\ln\frac{n+1}{n} = \ln \prod_{n=1}^\infty \left( \frac{n+1}{n} \right)^{(-1)^{n-1}} \\
= \lim_{N \to \infty} \ln \prod_{n=1}^N \frac{n^2}{(n - \frac{1}{2})(n + \frac{1}{2})} = \lim_{N \to \infty} \ln\frac{\pi \Gamma^2(N+1)}{2\Gamma(N + \frac{1}{2})\Gamma(N + \frac{3}{2})} = \ln \frac{\pi}{2}
$$
The sum therefore equals:
$$
S_1-S_2=\ln \frac{2\sqrt2}{\sqrt{\pi} \ \Gamma^2(\frac{3}{4})}
$$
as predicted.
A: Here is a fairly elementary proof of this result. I derive a closed form equivalent to the OP's.  
We begin with the following Lemma:

Lemma $1$: $\sum_{k=1}^p \ln(r*k+n) = \ln\left[r^n\left(\frac{n+r}{r}\right)_p\right] \quad \forall p\in\mathbb{N}$ 
Proof of Lemma $1$: $\sum _{k=1}^p\ln \left(rk+n\right) = \ln \left(\prod _{k=1}^p\left(rk+n\right)\right) = \ln \left(r^p\prod _{k=0}^{p-1}\left(k+\frac{n}{r}\right)\right) = \log\left[r^n\left(\frac{n+r}{r}\right)_p\right]$

We are now ready to prove the result
$$\begin{align} \sum_{n=1}^{p}(-1)^{n-1}&\left({\beta(n)\over n}-\ln{n+1\over n}\right)\\ &= \sum_{n=1}^{p}\left(-1\right)^{n-1}\left(\frac{1}{n}\sum _{k=0}^{\infty}\frac{\left(-1\right)^k}{\left(2k+1\right)^n}-\ln \left(\frac{n+1}{n}\right)\right)\\
&=\sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{n=1}^{p}(-1)^n\sum _{k=0}^{\infty}\frac{(-1)^k}{n\left(2k+1\right)^n}\\
&\stackrel{*}{=} \sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{k=0}^p\left(-1\right)^k\sum _{n=1}^\infty\frac{\left(-1\right)^n}{n\left(2k+1\right)^n}\\
&\stackrel{**}{=}\sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{k=0}^p\left(-1\right)^n\ln \left(\frac{2k+1}{2k+2}\right)\\
&=\sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{k=1}^p\left(-1\right)^k\ln \left(\frac{2k+1}{2k+2}\right)+\ln(2)\\
&=\sum _{k=1}^p\left(-1\right)^k\left[\ln \left(\frac{k+1}{k}\right)-\ln \left(\frac{2k+1}{2k+2}\right)\right]+\ln(2)\\
&=\sum _{k=1}^p\left[\ln \left(\frac{2k+1}{2k}\right)-\ln \left(\frac{4k+1}{4k+2}\right)-\ln \left(\frac{2k}{2k-1}\right)+\ln \left(\frac{4k-1}{4k}\right)\right]+\ln(2)\\
&= \sum _{k=1}^p\left(\ln \left(2k+1\right)-2\ln \left(2k\right)-\ln \left(4k+1\right)+\ln \left(4k+2\right)+\ln \left(2k-1\right)+\ln \left(4k-1\right)-\ln \left(4k\right)\right)+\ln(2)\\
&=\ln \left(2\left(3/2\right)_p\right)-2\ln \left(\left(1\right)_p\right)-\ln \left(4\left(5/4\right)_p\right)+\ln \left(16\left(3/2\right)_p\right)+\ln \left(\frac{1}{2}\left(1/2\right)_p\right)+\ln \left(\frac{1}{4}\left(3/4\right)_p\right)-\ln \left(\left(1\right)_p\right)+\ln(2)\\
&= \ln \left(\frac{2\cdot \left(1/2\right)_p\left(3/4\right)_p\left(3/2\right)_p\left(3/2\right)_p}{\left(1\right)_p\left(1\right)_p\left(1\right)_p\left(5/4\right)_p}\right)\\
&\stackrel{***}{=} \ln \left(\frac{2\cdot \Gamma\left(1/4\right)^2}{\pi ^2\sqrt{2\pi }}\right)+\ln \left(\frac{\Gamma\left(p+1/2\right)\Gamma\left(p+3/4\right)\Gamma\left(p+3/2\right)^2}{\Gamma\left(p+1\right)^3 \Gamma\left(p+5/4\right)}\right)
\end{align}$$
Luckily, as can be checked using Stirling's Formula, the second logarithm goes to $0$ as $p$ goes to infinity, i.e.
$$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right) = \ln \left(\frac{2\cdot \Gamma\left(1/4\right)^2}{\pi ^2\sqrt{2\pi }}\right) + \ln \left(\lim_{p \to \infty}\frac{\Gamma\left(p+1/2\right)\Gamma\left(p+3/4\right)\Gamma\left(p+3/2\right)^2}{\Gamma\left(p+1\right)^3 \Gamma\left(p+5/4\right)}\right)  = \color{red}{\ln \left(\frac{2\cdot \Gamma\left(1/4\right)^2}{\pi ^2\sqrt{2\pi }}\right)}$$

Footnotes:
$*\;\;\;\;\;\;$ Interchange double summations
 $**\;\;\;\,$ Taylor Series for Natural Logarithm
$***\;$ Gamma representation for Pochhammer Symbol
