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?
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) $