TITLE: Looking for the closed form of $\sum_{n=1}^{\infty}{\zeta(2n+1)\over (2n+1)2^{4n}}$
QUESTION [9 upvotes]: 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 $(3)?$

REPLY [5 votes]: 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

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

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