Any nice (not necessarily closed) forms for ${\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\eta(2n)}{n}}$? I've been playing around with series involving the eta function and I (think) managed to find a nice form for the series in the title (not a closed form, but a form with a very nice pattern). The derivation I did, however, was a bit tedious and not really rigorous (so I guess really it's possible my derivation is incorrect). I did, however, check numerically - and at least according to my computer - it seems to be correct.
Without spoiling what the one I found is; are there any other different nice forms of the above sum? Maybe one of the infinite series involving the zeta function could come in handy? (link: https://en.wikipedia.org/wiki/Riemann_zeta_function) Or turning it into a double sum?
Edit: the form I found is below, combined with another users answer you get an absolutely awesome result - check it out!
 A: Rewriting in terms of the zeta function, we have
$$
\sum_{n=1}^{\infty} \frac{\zeta(2n)(1-2^{1-2n})(-1)^{n+1}}{n}
$$
This lets us use the following identity:
$$
\sum_{n=1}^{\infty} \zeta(2n) x^{2n} = \frac{1-\pi x \cot(\pi x)}{2}
$$After several manipulations (reindexing, an integration, substitution), we are left with
$$
\log \left(\frac{1}{4} \pi  \sinh (\pi ) \text{csch}^2\left(\frac{\pi }{2}\right)\right)
$$

*

*https://mathworld.wolfram.com/RiemannZetaFunction.html
A: I took my own advice and also tried the double sum approach!
Expanding the ${\eta}$ function out, you get
$${\Rightarrow \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{2n}}=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}(-1)^{n+1}}{nk^{2n}}}$$
Swapping sums gives us
$${\Rightarrow \sum_{k=1}^{\infty}(-1)^{k+1}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\left(\frac{1}{k^2}\right)^{n}}{n}}$$
And since ${0 < \frac{1}{k^2}\leq 1}$;
$${\Rightarrow \sum_{k=1}^{\infty}(-1)^{k+1}\ln\left(1+\frac{1}{k^2}\right)=\ln\left(\left(\frac{1^2+1}{1^2}\right)\left(\frac{2^2}{2^2+1}\right)\left(\frac{3^2+1}{3^2}\right)\left(\frac{4^2}{4^2+1}\right)...\right)}$$
Which is also the form I originally found that I was talking about in the question (although I arrived here via a different method before). Altogether
$${\Rightarrow \sum_{n=1}^{\infty}\frac{(-1)^{n+1}\eta(2n)}{n}=\ln\left(\left(\frac{1^2+1}{1^2}\right)\left(\frac{2^2}{2^2+1}\right)\left(\frac{3^2+1}{3^2}\right)\left(\frac{4^2}{4^2+1}\right)...\right)}$$
Combined with @Integrand's awesome answer, you also get
$${\Rightarrow \left(\frac{1^2+1}{1^2}\right)\left(\frac{2^2}{2^2+1}\right)\left(\frac{3^2+1}{3^2}\right)\left(\frac{4^2}{4^2+1}\right)...=\frac{1}{4}\pi\sinh(\pi)\text{csch}^2\left(\frac{\pi}{2}\right)=\frac{\pi}{2}\coth\left(\frac{\pi}{2}\right)}$$
Which is so cool!!!
A: We begin with the infinite product representations of the hyperbolic sine and hyperbolic cosine functions as given by
$$\begin{align}
\sinh( z)&= z\prod_{n=1}^\infty \left(1+\frac{z^2}{(\pi n)^2}\right)\tag1\\\\
\cosh(z)&=\prod_{n=1}^\infty \left(1+\frac{z^2}{(\pi (n-1/2))^2}\right)\tag2
\end{align}$$


Using $(1)$ and $(2)$ and setting $z=\pi/2$ reveals
$$\begin{align}
\coth\left(\frac\pi2\right)&=\frac2\pi\prod_{n=1}^\infty \left(1+\frac1{(2n)^2}\right)^{-1}\left(1+\frac1{(2n-1)^2}\right)\tag3
\end{align}$$


Multiplying $(3)$ by $\pi/2$ and taking the logarithm, we obtain
$$\begin{align}
\log\left(\frac\pi2\coth\left(\frac\pi2\right)\right)&=\sum_{n=1}^\infty \left[\log\left(1+\frac1{(2n-1)^2}\right)-\log\left(1+\frac1{(2n)^2}\right)\right]\\\\
&=\sum_{n=1}^\infty (-1)^{n-1}\log\left(1+\frac{1}{n^2}\right)\\\\
&=\sum_{n=1}^\infty (-1)^{n-1}\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m}\frac1{n^{2m}}\\\\
&=\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^{2m}}\\\\
&=\sum_{m=1}^\infty \frac{(-1)^{m-1}\eta(2m)}{m}\tag4
\end{align}$$


Therefore, we find the coveted relationship
$$\bbox[5px,border:2px solid #C0A000]{\sum_{m=1}^\infty \frac{(-1)^{m-1}\eta(2m)}{m}=\log\left(\frac\pi2\coth\left(\frac\pi2\right)\right)}$$
and we are done!
