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.
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*}