Looking for the closed form of $\sum_{n=1}^{\infty}{\zeta(2n+1)\over (2n+1)2^{4n}}$

Question

We was able to determine $(1)$ to have this closed form

$$\ln(2)-\gamma=\sum_{n=1}^{\infty}{\zeta(2n+1)\over (2n+1)2^{2n}}\tag1$$

then we when on and try to evaluate $(2)$ and we only half of the closed form

$$2\ln(2)-\gamma-2X=\sum_{n=1}^{\infty}{\zeta(2n+1)\over (2n+1)2^{4n}}\tag2$$

Where $$X=\sum_{n=0}^{\infty}{\eta(2n+1)\over(2n+1)2^{2n+1}}\tag3$$

where $\eta$ is the Dirichlet eta function and $\gamma$ is Euler-Masheroni constant

How do we evaluate the closed form of $(3)?$

Answer 1

We have the following Lemma (a sketch of a proof can be found below)

$$ s(x)=\sum_{n\geq 1}(-x)^n \zeta(n+1)=-\gamma-\psi(1+x)\quad \color{red}{(I)} $$

where $\psi(z)=\frac{d\log(\Gamma(z))}{dz}$ is the digamma function and $\gamma$ is Euler's constant

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Integrating yields

$$ S(x)=\int dx s(x)=\sum_{n\geq1}\frac{\zeta(n+1)}{n+1}(-x)^{n+1}=-\gamma x-\log(\Gamma(1+x)) $$

Taking the odd part $$ S(x)-S(-x)=2\sum_{n\geq1}\frac{\zeta(2n+1)}{2n+1}x^{2n+1}=-2\gamma x-\log(\Gamma(1+x))+\log(\Gamma(1-x)) $$

Now let us put $x=\frac{1}{4}$ we get

$$ \sum_{n\geq1}\frac{\zeta(2n+1)}{2n+1}\frac1{4^{2n}}=-\gamma+2\log\left(\frac{\Gamma(3/4)}{\Gamma(5/4)}\right) $$

which is the sum of OP's interest

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We now proof $\color{red}{(I)}$:

Use the definition of the $\zeta$-function as a series and exchange the order of summation. Doing the first sum yields $S(x)=\sum_{k\geq1}\frac{1}{x+k}-\frac{1}k$ expressing this in terms of Digamma functions yields $\color{red}{(I)}$.

QED