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.
 A: 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
$$
A= \lim_{n \to \infty}\frac{1^12^2\cdots n^n}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}.
$$
A: 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 
$\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}\eta(n)=$ 
$\displaystyle = \frac{\eta(1)}{2}-\sum\limits_{n=2}^\infty\frac{(-1)^n}{n}\zeta(n)+2\sum\limits_{n=2}^\infty\frac{(-1)^n}{2^n n}\zeta(n)-\sum\limits_{n=2}^\infty\frac{(-1)^{n+1}}{n+1}\zeta(n)+4\sum\limits_{n=2}^\infty\frac{(-1)^{n+1}}{2^{n+1}(n+1)}\zeta(n)$
$\displaystyle =\frac{\eta(1)}{2}-( \ln Q_0(1)+\gamma )+2( \ln Q_0(\frac{1}{2})+\frac{\gamma}{2} )-( \ln Q_1(1)-\frac{\gamma}{2})+4(\ln Q_1(\frac{1}{2})-\frac{\gamma}{8})$
$\displaystyle =\frac{\ln 2}{2}+2 \ln\frac{\sqrt{\pi}}{2} -\ln\frac{\sqrt{2\pi}}{e} +4\ln(A_1^{\frac{3}{2}}2^{\frac{5}{24}}e^{-\frac{1}{2}})=\ln\frac{A^6\sqrt{\pi}}{2^{7/6}e}$
A: Another approach is to use a common integral representation of the Dirichlet eta function, along with an integral representation of the Hurwitz zeta function.
Specifically, we can use $$\eta(s)  = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}}{1+e^{x}} \, dx , \quad \text{Re}(s) >0,\tag{1}$$ and $$\zeta(s,z) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-zx}}{1-e^{-x}} \, dx \tag{2}, \quad (\text{Re}(s) >1, \, \text{Re}(z) >0). $$
From $(2)$ it follows that 
$$ \begin{align} \zeta(s)-\zeta \left(s, \frac{3}{2} \right) &= \frac{1}{\Gamma(s)} \int_{0}^{\infty}\frac{x^{s-1}}{1-e^{-x}} \left(e^{-x}-e^{-(3/2)x} \right) \\ &= \frac{2^{s}}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}}{1-e^{-2u}} \left(e^{-2u} - e^{-3u} \right) \, du  \\ &= \frac{2^{s}}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}e^{-2u}} {1+e^{-u}} \, du \\ &= \frac{2^{s}}{\Gamma(s)} \int_{0}^{\infty} \frac{u^{s-1}e^{-u}} {1+e^{u}} \, du, \quad \text{Re}(s) >0. \end{align}$$
And using $(1)$, we get $$ \begin{align} &\sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{n(n+1)} \, \eta(n) \\ &=  \sum_{n=1}^{\infty}\frac{(-1)^{n+1} }{n(n+1)}\frac{1}{\Gamma(n)} \int_{0}^{\infty} \frac{x^{n-1}}{1+e^{x}} \, dx \\ &= \int_{0}^{\infty} \frac{1}{1+e^{x}} \frac{1}{x^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!} \, \frac{x^{n+1}}{n+1} \, dx \\ &= \int_{0}^{\infty} \frac{1}{1+e^{x}} \frac{1}{x^{2}} \left(e^{-x}+x-1 \right) \, dx \\ &= \lim_{s \to -1^{+}} \int_{0}^{\infty} \frac{x^{s-1}}{1+e^{x}} \left(e^{-x}+x-1 \right) \, dx \\ &= \lim_{s \to -1^{+}} \left[2^{-s} \,  \Gamma(s) \left( \zeta(s) - \zeta \left(s, \frac{3}{2} \right)\right) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s) \right] \\&= \lim_{s \to -1^{+}} \left[2^{-s} \, \Gamma(s) \left(\zeta(s) - \zeta \left(s, \frac{1}{2} \right) +2^{s}\right) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s) \right] \tag{3} \\&= \lim_{s \to -1^{+}} \left[2^{-s} \, \Gamma(s) \left(\zeta(s) - \left(2^{s}-1\right)\zeta (s)  +2^{s}\right) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s) \right] \tag{4} \\ &= \lim_{s \to -1^{+}} \left[\left(2^{1-s}-1 \right) \Gamma(s) \zeta(s) + \Gamma(s) + \Gamma(s+1) \eta(s+1 ) - \Gamma(s) \eta(s)\right] \\&= \lim_{s \to -1^{+}} \left[ \Gamma(s) + \Gamma(s+1) \eta(s+1) -2\Gamma(s) \eta(s) \right] \tag{5}. \end{align}$$
Expanding about $s=-1$, we get 
$$ \begin{align} &\lim_{s \to -1^{+}} \left[ \Gamma(s) + \Gamma(s+1) \eta(s+1) -2\Gamma(s) \eta(s) \right] \\  &= \small \lim_{s \to -1^{+}} \Big[\left(- \frac{1}{s+1} + \gamma-1 + \mathcal{O}(s+1) \right) + \left(\frac{1}{s+1} - \gamma + \mathcal{O}(s+1) \right) \Big(\frac{1}{2} +\eta'(0)(s+1) \\ &+  \small \mathcal{O}\left( (s+1)^{2} \right) \Big) - \small  2 \left(- \frac{1}{s+1} + \gamma-1 + \mathcal{O}(s+1) \right) \left(\frac{1}{4} + \eta'(-1)(s+1) + \mathcal{O}\left((s+1)^{2} \right) \right) \Big] \\ &= \lim_{s \to -1^{+}} \left[\eta'(0) + 2 \eta'(-1) - \frac{1}{2} + \mathcal{O}\left((s+1) \right) \right]   \\&= \frac{1}{2} \, \ln \left(\frac{\pi}{2} \right) +2 \big( 4 \ln(2) \zeta(-1) -3 \zeta'(-1) \big) -\frac{1}{2} \tag{6} \\&= \frac{1}{2} \, \ln \left(\frac{\pi}{2} \right) +2 \left(-\frac{1}{3} \, \ln(2) -3 \left(\frac{1}{12} - \ln(A) \right)  \right) - \frac{1}{2} \tag{7} \\&= \ln \left(\frac{\sqrt{\pi} \, A^{6}}{ 2^{7/6} e} \right) . \end{align}$$

$(3)$ $\zeta(s,a) = \zeta(s,a+1) + a^{-s}$
$(4)$ http://mathworld.wolfram.com/HurwitzZetaFunction.html (11)
$(5)$ $\eta(s) = (1-2^{1-s}) \zeta(s)$
$(6)$ http://mathworld.wolfram.com/DirichletEtaFunction.html (11)
$(7)$ https://en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant

To expand $\Gamma(s+1)$ in a Laurent series about $s=-1$, use the identity $\Gamma(s+1) = \frac{\Gamma(s+2)}{s+1}$.
