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

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

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?

Solutions Collecting From Web of "$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}?$"

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.

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$\substack{\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n – 1}\bracks{{\beta\pars{n} \over n} -\ln\pars{n + 1 \over n}} =
\ln\pars{\root{2 \over \pi}\,{2 \over \Gamma^{\,2}\pars{3/4}}}:\ {\large ?}.}
\\[3mm]
\ds{\beta:\texttt{Dirichlet Beta Function}.}}$

Lets
\begin{equation}
\sum_{n = 1}^{\infty}\pars{-1}^{n – 1}\bracks{{\beta\pars{n} \over n} -\ln\pars{n + 1 \over n}} = \mc{S}_{1} – \mc{S}_{2}\,,\quad
\left\{\begin{array}{l}
\ds{\mc{S}_{1} \equiv
\sum_{n = 1}^{\infty}\pars{-1}^{n – 1}\,{\beta\pars{n} \over n}}
\\[2mm]
\ds{\mc{S}_{2} \equiv
\sum_{n = 1}^{\infty}\pars{-1}^{n – 1}\,\ln\pars{n + 1 \over n}}
\end{array}\right.\label{1}\tag{1}
\end{equation}


$\ds{\Large\mc{S}_{1}:\ ?.}$ With the $\ds{\beta}$ Integral Representation:
\begin{align}
\mc{S}_{1} & \equiv
\sum_{n = 1}^{\infty}\pars{-1}^{n – 1}\,{1 \over n}\
\overbrace{{1 \over \Gamma\pars{n}}\int_{0}^{\infty}
{x^{n – 1}\expo{-x} \over 1 + \expo{-2x}}\,\dd x}^{\ds{\beta\pars{n}}}\ =\
-\int_{0}^{\infty}
{\bracks{\sum_{n = 1}^{\infty}\pars{-x}^{n}/n!}\expo{-x} \over 1 + \expo{-2x}}\,{\dd x \over x}
\\[5mm] & =
-\int_{0}^{\infty}
{\pars{\expo{-x} – 1}\expo{-x} \over 1 + \expo{-2x}}\,{\dd x \over x}
\,\,\,\stackrel{\exp\pars{-2x}\ \mapsto\ x}{=}\,\,\,
-\int_{1}^{0}{x – x^{1/2} \over 1 + x}\,{\dd x/\pars{-2x} \over \ln\pars{x}/\pars{-2}} \\[5mm] & =
\int_{0}^{1}{x^{-1/2} – 1 \over 1 – x^{2}}\,{x – 1 \over \ln\pars{x}}\,\dd x =
\int_{0}^{1}{x^{-1/2} – 1 \over 1 – x^{2}}\
\overbrace{\int_{0}^{1}x^{t}\,\dd t}^{\ds{x – 1 \over \ln\pars{x}}}\
\,\dd x
\\[5mm] & =
\int_{0}^{1}\int_{0}^{1}{x^{t – 1/2} – x^{t} \over 1 – x^{2}}\,\dd x\,\dd t
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 2}\int_{0}^{1}\int_{0}^{1}{x^{t/2 – 3/4} – x^{t/2 – 1/2} \over
1 – x}\,\dd x\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\bracks{%
\int_{0}^{1}{1 – x^{t/2 – 1/2} \over 1 – x}\,\dd x –
\int_{0}^{1}\int_{0}^{1}{1 – x^{t/2 – 3/4} \over 1 – x}\,\dd x}\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}
\bracks{\Psi\pars{{t \over 2} + {1 \over 2}} –
\Psi\pars{{t \over 2} + {1 \over 4}}}\dd t
\qquad\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] & =
\left.
\ln\pars{\Gamma\pars{t/2 + 1/2} \over \Gamma\pars{t/2 + 1/4}}
\right\vert_{\ 0}^{\ 1} =
\ln\pars{{\Gamma\pars{1} \over \Gamma\pars{3/4}}\,{\Gamma\pars{1/4} \over \Gamma\pars{1/2}}}
\\[5mm] & =
\ln\pars{{1 \over \Gamma^{\,2}\pars{3/4}}\,{\Gamma\pars{3/4}\Gamma\pars{1/4} \over \root{\pi}}}\quad
\pars{~\substack{\mbox{Note that}\\[2mm]
\ds{\Gamma\pars{1 \over 2} = \root{\pi}\,,\ \Gamma\pars{1} = 1}}~}
\\[5mm] & =
\ln\pars{{1 \over \Gamma^{\,2}\pars{3/4}\root{\pi}}
\,{\pi \over \sin\pars{\pi/4}}}\qquad\pars{~Euler\ Reflection\ Formula~}
\\[5mm] & \implies
\bbx{\mc{S}_{1} \equiv
\sum_{n = 1}^{\infty}\pars{-1}^{n – 1}\,{\beta\pars{n} \over n} =
\ln\pars{\root{2\pi} \over \Gamma^{\,2}\pars{3/4}}}\label{2}\tag{2}
\end{align}


$\ds{\Large\mc{S}_{2}:\ ?.}$
\begin{align}
\mc{S}_{2} & \equiv
\sum_{n = 1}^{\infty}\pars{-1}^{n – 1}\,\ln\pars{n + 1 \over n} =
\sum_{n = 0}^{\infty}\pars{-1}^{n}\int_{0}^{1}{\dd t \over t + n + 1} =
\int_{0}^{1}\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n + t + 1}\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\sum_{n = 0}^{\infty}
\bracks{{1 \over n + \pars{t + 1}/2} – {1 \over n + t/2 + 1}}\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\bracks{%
\Psi\pars{{t \over 2} + 1} – \Psi\pars{t + 1 \over 2}}\,\dd t =
\left.\ln\pars{\Gamma\pars{t/2 + 1} \over \Gamma\pars{t/2 + 1/2}}
\right\vert_{\ 0}^{\ 1}
\\[5mm] & =
\ln\pars{{\Gamma\pars{3/2} \over \Gamma\pars{1}}\,{\Gamma\pars{1/2} \over \Gamma\pars{1}}}
=
\ln\pars{{1 \over 2}\,\Gamma^{\,2}\pars{1 \over 2}}
\qquad\pars{~\Gamma-Recursive Property~}
\\[5mm] & \implies
\bbx{\mc{S}_{2} \equiv
\sum_{n = 1}^{\infty}\pars{-1}^{n – 1}\,\ln\pars{n + 1 \over n} =
\ln\pars{\pi \over 2}}\label{3}\tag{3}
\end{align}


With \eqref{1}, \eqref{2} and \eqref{3}:
$$
\bbx{\sum_{n = 1}^{\infty}\pars{-1}^{n – 1}\bracks{{\beta\pars{n} \over n} -\ln\pars{n + 1 \over n}} =
\ln\pars{\root{2 \over \pi}\,{2 \over \Gamma^{\,2}\pars{3/4}}}}
$$

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