Sum $\sum _{n=1}^{\infty \:}\frac{\left(-1\right)^{n-1}}{\left(\left(2n-1\right)^2\right)}$ How to find sum $\sum\limits _{n=1}^{\infty \:}\frac{\left(-1\right)^{n-1}}{\left(2n-1\right)^2}$ from sum $\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}$?
The first sum I found by integration of Fourier series of function $f(x)=x$ for $-\pi\leq x\leq \pi$, but I still don't know how to find the other sum.
Any hint or help is welcome. Thanks in advance.
 A: The way to understand both is to consider $\sum_{n=1}^\infty \frac{z^n}{n^2}$ and
$\sum_{n=1}^\infty \frac{(-z)^n}{n^2}$.
A: You want 
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n-1)^2} = G =0.915 \cdots
\end{eqnarray*}
The Catalan constant.
Of course you know Euler's solution to the Basel problem
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{ \pi^2}{6}
\end{eqnarray*}
and from this it is easy to calculate the sum of the reciprocals of squares of ... (positive ) even numbers , odd numbers ... and if you subtract these you will get the alternating result that you state in your question
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} = \frac{ \pi^2}{12}.
\end{eqnarray*}
So the next obvious question is the one you are asking ... alternating odd numbers ... and it would seem that this cannot be simplified & it is called the Catalan constant.
Note also that
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{ \pi^4}{90}.
\end{eqnarray*}
So one might reasonably suppose that 
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{n^3} = r \pi^3
\end{eqnarray*}
where $r$ is a rational value. But it turns out not to be the case ... & this is called Aprey's constant ... check it out
https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant
Even more wierd ... the alternating sum of reciprocals of (positive) odd numbers cubed can be simplified 
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n-1)^3} = \frac{ \pi^3}{32}.
\end{eqnarray*}
