Prove that $\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum\limits_{n=1}^{\infty}{(-1)^{n+1}\over n(n+1)}\eta(n)$

Show that

$$\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n(n+1)}\eta(n)$$

(where $\eta(n)$ is the Dirichlet eta function, and A is the Glaisher-Kinkelin constant).

I try:

$$\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n}\eta(n)-\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n)$$

We use this series $$\ln{{\pi\over 2}}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n}\eta(n)$$ to simplify to

$$-\ln{\left({A^6\over 2^{1\over6}\sqrt{\pi}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n)$$

We have

$$\ln{\left[\prod_{k=1}^{\infty}\left({k\over k+1}\right)^{(-1)^{k+1}}\right]}=\sum_{n=1}^{\infty}{(-1)^n\eta(n)\over n}$$

We can't use this to apply on the above series.

I just wonder, is there an infinite product for

$$\ln{\left[F(k)\right]}=\sum_{n=1}^{\infty}{(-1)^n\eta(n)\over 1+n}$$

Can anyone please give a hand here? Thank you.

Hint. One may write $$ \begin{align} \sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n)&=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\eta(n) \\&=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\left(1+\sum_{k=2}^{\infty}\frac{(-1)^{k-1}}{k^n}\right) \\&=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}+\sum_{n=1}^{\infty}{(-1)^{n+1}\over n+1}\sum_{k=2}^{\infty}\frac{(-1)^{k-1}}{k^n} \\&=1-\ln 2+\sum_{k=2}^{\infty}(-1)^{k-1}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{(n+1)k^n} \\&=1-\ln 2+\sum_{k=2}^{\infty}(-1)^{k-1}\left[1-k\ln\left(1+\frac1k \right) \right] \\&=\sum_{k=1}^{\infty}(-1)^{k}\left[k\ln\left(1+\frac1k \right)-1 \right] \tag1 \end{align} $$ since, by a Taylor series expansion, as $k\to \infty$, one has $$ (-1)^{k}\left[k\ln\left(1+\frac1k \right)-1 \right]=\frac{(-1)^{k}}{2k}+O\left(\frac{1}{k^2}\right) $$ the series in $(1)$ is thus convergent, it is then sufficient to consider its partial sum $$ \begin{align} S_{2n}&=\sum_{k=1}^{2n}(-1)^{k}\left[k\ln\left(1+\frac1k \right)-1 \right] \\&=\sum_{k=1}^{2n}(-1)^{k}k\ln\left(k+1\right)-\sum_{k=1}^{2n}(-1)^{k}k\ln k-\sum_{k=1}^{2n}(-1)^{k} \\&=\sum_{k=2}^{2n+1}(-1)^{k-1}(k-1)\ln k-\sum_{k=1}^{2n}(-1)^{k}k\ln k \\&=2n\ln (2n+1)-2\sum_{k=1}^{2n}(-1)^{k}k\ln k+\sum_{k=1}^{2n}(-1)^{k}\ln k \tag2 \end{align} $$ the latter sum is obtained via the Stirling formula $$ \sum_{k=1}^{2n}(-1)^{k}\ln k=\frac{\ln \pi}2+\frac{\ln n}2+O\left(\frac1n \right) $$ and one may rearrange the other sum $$ \begin{align} &\sum_{k=1}^{2n}(-1)^{k-1}k\ln k \\&=\sum_{k=1}^{2n}k\ln k-2\sum_{k=1}^{n}2k\ln(2k) \\&=\sum_{k=1}^{2n}k\ln k-4\sum_{k=1}^{n}k\ln k-4\sum_{k=1}^{n}k\ln 2 \\&=\ln \frac{1^12^2\cdots (2n)^{2n}}{(2n)^{2n^2+n+1/12}e^{-n^2}}-4\ln \frac{1^12^2\cdots n^n}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}-\left(n-\frac1{12}\right)\ln 2-\left(n+\frac14\right)\ln n \end{align} $$ giving in $(2)$ $$ S_{2n}=\frac{\ln \pi}2+\frac{\ln 2}6+1-6\ln A+O\left(\frac1n \right) $$ as expected, since $$ A= \lim_{n \to \infty}\frac{1^12^2\cdots n^n}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}. $$

Using the generalisation $Q_m(x)$ of the Gamma function in

https://www.fernuni-hagen.de/analysis/download/bachelorarbeit_aschauer.pdf

with $\enspace\displaystyle \ln Q_m(x)=\frac{(-x)^{m+1}}{m+1}\gamma +\sum\limits_{n=2}^\infty\frac{(-x)^{m+n}}{m+n}\zeta(n)\enspace$ on page $13$, $(4.2)$, with

$Q_0(1)=\Gamma(2)=1\enspace$ and $\enspace\displaystyle Q_0(\frac{1}{2})=\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}\enspace$ and with the calculated values

$\displaystyle Q_1:=Q_1(1)=\frac{\sqrt{2\pi}}{e}\enspace$ and $\enspace\displaystyle Q_1(\frac{1}{2})= A_1^{\frac{3}{2}}2^{\frac{5}{24}}e^{-\frac{1}{2}} \enspace$ on page $38$ $\enspace$ where

$A_1\equiv A\enspace$ (Glaisher Kinkelin constant) we get

$\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}\eta(n)=$

$\displaystyle = \frac{\eta(1)}{2}-\sum\limits_{n=2}^\infty\frac{(-1)^n}{n}\zeta(n)+2\sum\limits_{n=2}^\infty\frac{(-1)^n}{2^n n}\zeta(n)-\sum\limits_{n=2}^\infty\frac{(-1)^{n+1}}{n+1}\zeta(n)+4\sum\limits_{n=2}^\infty\frac{(-1)^{n+1}}{2^{n+1}(n+1)}\zeta(n)$

$\displaystyle =\frac{\eta(1)}{2}-( \ln Q_0(1)+\gamma )+2( \ln Q_0(\frac{1}{2})+\frac{\gamma}{2} )-( \ln Q_1(1)-\frac{\gamma}{2})+4(\ln Q_1(\frac{1}{2})-\frac{\gamma}{8})$

$\displaystyle =\frac{\ln 2}{2}+2 \ln\frac{\sqrt{\pi}}{2} -\ln\frac{\sqrt{2\pi}}{e} +4\ln(A_1^{\frac{3}{2}}2^{\frac{5}{24}}e^{-\frac{1}{2}})=\ln\frac{A^6\sqrt{\pi}}{2^{7/6}e}$