Any nice (not necessarily closed) forms for ${\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\eta(2n)}{n}}$?

Question

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!

Answer 1

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

Answer 2

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!!!

Answer 3

We begin with the infinite product representations of the hyperbolic sine and hyperbolic cosine functions as given by

$$\begin{align} \sinh( z)&= z\prod_{n=1}^\infty \left(1+\frac{z^2}{(\pi n)^2}\right)\tag1\\\\ \cosh(z)&=\prod_{n=1}^\infty \left(1+\frac{z^2}{(\pi (n-1/2))^2}\right)\tag2 \end{align}$$

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Using $(1)$ and $(2)$ and setting $z=\pi/2$ reveals

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

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Multiplying $(3)$ by $\pi/2$ and taking the logarithm, we obtain

$$\begin{align} \log\left(\frac\pi2\coth\left(\frac\pi2\right)\right)&=\sum_{n=1}^\infty \left[\log\left(1+\frac1{(2n-1)^2}\right)-\log\left(1+\frac1{(2n)^2}\right)\right]\\\\ &=\sum_{n=1}^\infty (-1)^{n-1}\log\left(1+\frac{1}{n^2}\right)\\\\ &=\sum_{n=1}^\infty (-1)^{n-1}\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m}\frac1{n^{2m}}\\\\ &=\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^{2m}}\\\\ &=\sum_{m=1}^\infty \frac{(-1)^{m-1}\eta(2m)}{m}\tag4 \end{align}$$

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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!