While playing around with the problems here 1 and here 2, I came to found the double sum identity for Catalan constant, G by making few change in those problems. ie
$$ G= \sum_{m=1}^{\infty}\sum_{n=0}^{\infty} \frac{1}{(2m+n)^2}\left((-1)^m-(-1)^{m+n}+(-1)^n\left(1+\frac{2m}{(m+n)}+\frac{m^2}{(m+n)^2}\right)\right)$$
To prove this I had to evaluate 3 different summation ( just apply the linearity) which makes work quite tedious and length for me. That is I had to evaluate the following summation (simplified form)
$$\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}\left(\frac{(-1)^m}{(2m+n)^2}-\frac{(-1)^{m+n}}{(2m+n)^2}+\frac{(-1)^n}{(m+n)^2}\right)\tag{1}$$
Evaluation of initial sum $$\sum_{m=1}^{\infty} \sum_{n=0}^{\infty} \frac{(-1)^m}{(2m+n)^2}=\sum_{m=1}^{\infty}(-1)^m\sum_{n=0}^{\infty}\frac{1}{(2m+n)^2}$$
If we write the inner sum in series form we notice that $$\sum_{n=0}^{\infty} \dfrac{1}{(2m+n)^2}= \underbrace{\dfrac{1}{4}\sum_{n=0}^{\infty} \dfrac{1}{(m+n)^2} }_{\text{}S_1}+\underbrace {\sum_{n=1,3,5\cdots}^{\infty} \dfrac{1}{(2m+n)^2}}_{\text{} S_2}$$ and hence have $$\sum_{m=1}^{\infty} \sum_{n=0}^{\infty} \dfrac{(-1)^m}{(2m+n)^2}=\sum_{m=1}^{\infty}(-1)^m \underbrace{\dfrac{1}{4}\sum_{n=0}^{\infty} \dfrac{1}{(m+n)^2} }_{\text{}S_1}+\sum_{m=1}^{\infty} (-1)^m\underbrace {\sum_{n=1,3,5\cdots }^{\infty} \dfrac{1}{(2m+n)^2}}_{\text{} S_2}$$ we now evaluate the $S_1$ and $S_2$ separately as follows $$\begin{align}4S_1 =\sum_{m=1}^{\infty} \sum_{n=0}^{\infty}\dfrac{(-1)^m}{(m+n)^2}=\sum_{m=1}^{\infty} (-1)^m\left(\dfrac{1}{m^2}+\dfrac{1}{(m+1)^2}\right)\\ +\sum_{m=1}^{\infty}(-1)^m \left(\dfrac{1}{(m+2)^2}+\dfrac{1}{(m+3)^2}\right)+\cdots \\=\underbrace{\sum_{m=1}^{\infty} \left(\dfrac{1}{(m+1)^2}-\dfrac{1}{m^2}\right)}_{\text {Telescopes}}+\underbrace{\sum_{m=1}^{\infty} \left(\dfrac{1}{(m+3)^2}-\dfrac{1}{(m+2)^2}\right)}_{\text {Telescopes}}+\cdots \\ =-\left(\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \right) =-\sum_{m=1}^{\infty}\dfrac{1}{(2m-1)^2}\\=-\left(\sum_{m=1}^{\infty}\dfrac{1}{m^2}-\dfrac{1}{4}\sum_{m=1}^{\infty}\dfrac{1}{m^2}\right)=-\dfrac{\pi^2}{8}\end{align}$$ therefore we have $S_1=-\frac{\pi^2}{32}$ and we are left to evaluate $$ \begin{align} S_2= \sum_{m=1}^{\infty}\sum_{n=1,3,5\cdots }^{\infty}\dfrac{(-1)^m}{(2m+n)^2}=\sum_{m=1}^{\infty} (-1)^m\left(\dfrac{1}{(2m+1)^2}+\dfrac{1}{(2m+3)^2}\right)\\ +\sum_{m=1}^{\infty}(-1)^m \left(\dfrac{1}{(2m+5)^2}+\dfrac{1}{(2m+7)^2}\right)+\cdots \\=\underbrace{\sum_{m=1}^{\infty} \left(\dfrac{1}{(2m+3)^2}-\dfrac{1}{(2m+1)^2}\right)}_{\text {Telescopes}}+\underbrace{\sum_{m=1}^{\infty} \left(\dfrac{1}{(2m+7)^2}-\dfrac{1}{(2m+5)^2}\right)}_{\text {Telescopes}}+\cdots\\=-\left(\dfrac{1}{3^2}+\dfrac{1}{7^2}+\dfrac{1}{11^2}+\cdots\right)=-\sum_{m=0}^{\infty}\dfrac{1}{(4m+3)^2}\\=-\dfrac{1}{4^2} \psi^1\left(\dfrac{3}{4}\right) =-\dfrac{1}{16}\left(\pi^2-8G\right)\end{align}$$ or $$\begin{align} \sum_{m=0}^{\infty}\dfrac{1}{(4m+3)^2} = \sum_{m=1}^{\infty}\dfrac{1}{m^2}-\left(\sum_{m=1}^{\infty}\dfrac{1}{(2m)^2}+\sum_{m=1}^{\infty}\dfrac{1}{(4m-3)^2}\right)\\= \dfrac{\pi^2}{8}- \left(\sum_{m=0}^{\infty}\dfrac{(-1)^m}{(2m-1)^2}+\sum_{m=0}^{\infty}\dfrac{1}{(4m+3)^2}\right)\\\implies \sum_{m=0}^{\infty}\dfrac{1}{(4m+3)^2} =\dfrac{\pi^2}{2\times 8}-\dfrac{G}{2} \end{align}$$ thus $$\sum_{m=1}^{\infty} \sum_{n=0}^{\infty} \dfrac{(-1)^m}{(2m+n)^2}=-\dfrac{\pi^2}{32}+\dfrac{G}{2}-\dfrac{\pi^2}{16} =\dfrac{G}{2} -\frac{3}{32}\pi^2$$ Similarly, evaluating the middle and last summation i came up with the closed form $\displaystyle -\frac{G}{2}+\frac{\pi^2}{32}$ and $\displaystyle \frac{\pi^2}{8}$ giving us $$\frac{G}{2}-\frac{3}{32}\pi^2+\frac{G}{2}-\frac{\pi^2}{32}+\frac{\pi^2}{8}=G$$
Now my query is, How to evaluate/ relate the double summation without evaluating the each individual summation?
One of the way I had solved it using integration trickes which is solved here by Sergio Esteban, Argentina.
Thank you !!
A solution using $1/a^2=\int_0^1 x^{a-1}l(x)\,dx$ for $a>0$, where $l(x)=-\log x$ for brevity. \begin{align*} (1)\quad&=\sum_{m=1}^\infty(-1)^m\sum_{n=0}^\infty\frac{1-(-1)^n}{(2m+n)^2}+\sum_{m=1}^\infty\sum_{n=0}^\infty\frac{(-1)^n}{(m+n)^2} \\&=\sum_{m=1}^\infty\sum_{n=0}^\infty\left(\frac{2(-1)^m}{(2m+2n+1)^2}+\frac{(-1)^n}{(m+n)^2}\right) \\&=\int_0^1\left(\sum_{m=1}^\infty\sum_{n=0}^\infty\big(2(-1)^m x^{2m+2n}+(-1)^n x^{m+n-1}\big)\right)l(x)\,dx \\&=\int_0^1\left(2\frac{-x^2}{1+x^2}\frac{1}{1-x^2}+\frac{1}{1-x}\frac{1}{1+x}\right)l(x)\,dx=\color{blue}{\int_0^1\frac{l(x)}{1+x^2}\,dx} \\&=\int_0^1\left(\sum_{n=0}^\infty(-1)^n x^{2n}\right)l(x)\,dx=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}=G. \end{align*}