I've been playing with series involving odd values of the zeta function. Some time ago I found the following closed form $$ \sum_{k=1}^{\infty}\frac{\eta(2k+1)}{2^{2k+1}}=\frac{1}{2}-\ln(2) $$ and recently I stumbled upon this one $$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{\zeta(2n+1)}{2^{2n+1}} $$ but had no luck, yet, finding a nice closed form. When I say "nice", I mean in terms of $ln$, $\pi$, etc.
So, maybe someone here can find and share the hypothetical nice closed form.
Thanks.
On
We can conclude the Simple Art's answer observing that $$\sum_{k\geq1}\frac{1}{k\left(4k^{2}+1\right)}=\sum_{k\geq1}\frac{1}{k\left(2k+i\right)\left(2k-i\right)} $$ $$\frac{i}{4}\sum_{k\geq1}\frac{1}{k\left(k+\frac{i}{2}\right)}-\frac{i}{4}\sum_{k\geq1}\frac{1}{k\left(k-\frac{i}{2}\right)} $$ $$=\frac{1}{2}\left(2\gamma+\psi^{\left(0\right)}\left(1+\frac{i}{2}\right)+\psi^{\left(0\right)}\left(1-\frac{i}{2}\right)\right) $$ hence $$\sum_{n\geq1}\left(-1\right)^{n+1}\frac{\zeta\left(2n+2\right)}{2^{2n+1}}=-\frac{1}{4}\left(2\gamma+\psi^{\left(0\right)}\left(1+\frac{i}{2}\right)+\psi^{\left(0\right)}\left(1-\frac{i}{2}\right)\right). $$
This is not a complete answer.
$$\sum_{n=1}^\infty(-1)^{n+1}\frac{\zeta(2n+1)}{2^{2n+1}}=\sum_{n=1}^\infty(-1)^{n+1}\frac{\sum_{k=1}^\infty\frac1{k^{2n+1}}}{2^{2n+1}}$$
$$=\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+1}\frac{\frac1{k^{2n+1}}}{2^{2n+1}}$$
$$=\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+1}\frac1{(2k)^{2n+1}}$$
$$=\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+1}\frac1{2k(4k^2)^n}$$
$$=\sum_{k=1}^\infty\frac{-1}{2k}\sum_{n=1}^\infty(-1)^n\left(\frac1{4k^2}\right)^n$$
$$=\sum_{k=1}^\infty\frac{-1}{2k}\frac1{1+4k^2}$$
$$=-\frac12\sum_{k=1}^\infty\frac1{k(1+4k^2)}$$
According to Wolframalpha
$$\sum_{n=1}^\infty(-1)^{n+1}\frac{\zeta(2n+1)}{2^{2n+1}}=\frac12\left(2\gamma+\psi^{(0)}(1-\frac i2)+\psi^{(0)}(1+\frac i2)\right)$$
And looking at the integral representation,
$$=\gamma+\Re\psi^{(0)}(1+\frac i2)$$