Similar sums $\sum _{n=1}^\infty \frac{(-1)^{n-1}}{n^3}$ and $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi ^3}{32}$ Is it possible to find $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}$ if we know that $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{32}$?
Any help is welcome.
Thanks in advance.
 A: The short answer is no, but nevertheless they are some how related as we will shortly see below.
Lets start by the first series which is an instance of the Dirichlet eta function $\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$, which is related to the Riemann Zeta function by the equation
\begin{align*}
\eta(s)=(1-2^{1-s})\zeta(s) \tag{1}
\end{align*}
In your case $s=3$, therefore,  by $(1)$ we obtain
\begin{align*}
\sum_{n=1}^\infty  \frac{(-1)^{n-1}}{n^3}=\frac34\zeta(3) 
\end{align*}
The second series is an instance of the Dirichlet Beta function which is defined by
\begin{align*}
\beta(s)=\sum_{n=0}^\infty  \frac{(-1)^{n}}{(2n+1)^s}
\end{align*}
One relation between these two functions can be established by yet another function, the polylogarithm given by
\begin{align*}
\operatorname{Li}_s(x)=\sum_{n=1}^\infty  \frac{x^n}{n^s} \tag{2}
\end{align*}
Setting $x=\sqrt{-1}=i$ in $(2)$ we obtain
\begin{align*}
\operatorname{Li}_s(i)&=\sum_{n=1}^\infty  \frac{i^n}{n^s} \\
&=\frac{i}{1^s}-\frac{1}{2^s}-\frac{i}{3^s}+\frac{1}{4^s}+ \cdots\\
&=i\sum_{n=0}^\infty \frac{(-1)^s}{(2n+1)^s}+\sum_{n=1}^\infty \frac{(-1)^n}{(2n)^s}\\
&=i\beta(s)-\frac{1}{2^s}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}\\
&=i\beta(s)-2^{-s}\eta(s)\\
 \end{align*}
For $s=3$ we obtain
\begin{align*}
\operatorname{Li}_3(i)&=i\beta(3)-\frac18\eta(3)\\
&=i \frac{\pi^3}{32}-\frac{3}{32}\zeta(3)
 \end{align*}
One interesting fact to note is that the Riemann zeta function and consequently the Dirichlet eta function can be expressed in closed form for $n$ even, and the Dirichlet Beta function can be expressed in closed form for odd indices.
Hope this somehow helps you.
