Closed form for $\sum\limits_{n=1}^{\infty}\frac{O_{n}^{(p)}}{(2n-1)^{q}}$, with $O_{n}^{(s)}=\sum\limits_{k=1}^n\frac1{(2k-1)^{s}}$ Consider the sum 

$$\sum_{n=1}^{\infty}\frac{O_{n}^{(p)}}{(2n-1)^{q}}\text{, with }O_{n}^{(s)}=1+\frac{1}{3^{s}}+\dots+\frac{1}{(2n-1)^{s}}$$

My question is: if there exists some general theorems that allow to represent this sums in terms of the Riemann Zeta function, like Borwein-Borwein-Girgensohn theorem for Euler sums?
 A: Changing your notation a bit
and playing around.
If
$O_{n}(s)
=1+\frac{1}{3^{s}}+\dots+\frac{1}{(2n-1)^{s}}
=\sum_{k=1}^n \frac1{(2k-1)^s}
$,
$O(s)
=O_{\infty}(s)
$,
and
$s(p, q)
=\sum_{n=1}^{\infty}\dfrac{O_{n}(p)}{(2n-1)^{q}}
$
then
$s(p, q)+s(q, p)
=O(p)O(q)+O(p+q)
$.
Proof.
$\begin{array}\\
s(p, q)
&=\sum_{n=1}^{\infty}\dfrac{O_{n}(p)}{(2n-1)^{q}}\\
&=\sum_{n=1}^{\infty}\dfrac{\sum_{k=1}^n \frac1{(2k-1)^p}}{(2n-1)^{q}}\\
&=\sum_{n=1}^{\infty}\sum_{k=1}^n \dfrac1{(2k-1)^p}\dfrac{1}{(2n-1)^{q}}\\
&=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty} \dfrac1{(2k-1)^p}\dfrac{1}{(2n-1)^{q}}\\
&=\sum_{k=1}^{\infty}\dfrac1{(2k-1)^p}\sum_{n=k}^{\infty} \dfrac{1}{(2n-1)^{q}}\\
&=\sum_{k=1}^{\infty}\dfrac1{(2k-1)^p}\left(\sum_{n=1}^{\infty} \dfrac{1}{(2n-1)^{q}}-\sum_{n=1}^{k-1} \dfrac{1}{(2n-1)^{q}}\right)\\
&=\sum_{k=1}^{\infty}\dfrac1{(2k-1)^p}\left(O(q)-O_{k-1}(q)\right)\\
&=O(p)O(q)-\sum_{k=1}^{\infty}\dfrac1{(2k-1)^p}O_{k-1}(q)\\
&=O(p)O(q)-\sum_{k=1}^{\infty}\dfrac1{(2k-1)^p}(O_{k}(q)-\frac1{(2k-1)^q})\\
&=O(p)O(q)-\sum_{k=1}^{\infty}\dfrac1{(2k-1)^p}O_{k}(q)+\sum_{k=1}^{\infty}\dfrac1{(2k-1)^p}\dfrac1{(2k-1)^q}\\
&=O(p)O(q)-s(q, p)+\sum_{k=1}^{\infty}\dfrac1{(2k-1)^{p+q}}\\
&=O(p)O(q)-s(q, p)+O(p+q)
\\
\end{array}
$
Some additional 
definitions and stuff
which might be useful.
$Z_{n}(s)
=\sum_{k=1}^n \frac1{k^s}
$
$E_{n}(s)
=\sum_{k=1}^n \frac1{(2k)^s}
=\frac1{2^s}\sum_{k=1}^n \frac1{k^s}
=\frac1{2^s}Z_{n}(s)
$
$O_{n}(s)+E_{n}(s)
=\sum_{k=1}^{2n} \frac1{k^s}
=Z_{2n}(s)
$
so
$O_{n}(s)
=Z_{2n}(s)-E_{n}(s)
=Z_{2n}(s)-\frac1{2^s}Z_{n}(s)
$
$O(s)
=O_{\infty}(s)
$,
$E(s)
=E_{\infty}(s)
$,
$Z(s)
=\zeta(s)
=Z_{\infty}(s)
$.
$O(s)
=Z(s)-E(s)
=Z(s)-\frac1{2^s}Z(s)
=(1-2^{-s})Z(s)
$
