How to evaluate this double infinite sum (Catalan number) Let  $C_n = \dfrac{1}{n+1}\binom{2n}{n}$.
Is it possible to find the exact value of this infinite sum ?
$$\sum_{n=1}^\infty \sum_{k=n}^\infty \frac{\left(C_{n+1}-2C_n\right)\left(C_{k+1}-C_k\right)}{4^{n+k}}$$
In general, is it possible to evaluate 
$$\sum_{n=1}^\infty \sum_{k=n}^\infty \frac{\left(C_{n+1}-2C_n\right)\left(C_{k+1}-C_k\right)}{x^{n+k}}$$
in term of $x$?
Thanks in advances.
 A: We have:
$$ C_{k+1}-C_k = \frac{3k\, 4^k\, \Gamma\left(k+\frac{1}{2}\right)}{\sqrt{\pi}\,\Gamma(k+3)}\tag{1} $$
hence:
$$ \sum_{k\geq n}\frac{C_{k+1}-C_k}{4^k} =\frac{(6n+2)\,\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}\,\Gamma(n+2)} \tag{2}$$
and:
$$\begin{eqnarray*} \sum_{n\geq 1}\frac{C_{n+1}-2C_n}{4^n}\sum_{k\geq n}\frac{C_{k+1}-C_k}{4^k} &=& \frac{4}{\pi}\sum_{n\geq 1}\frac{(n-1)\,\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+3)}\cdot \frac{(3n+1)\,\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+2)}\\&=&\frac{4}{\pi}\sum_{n\geq 1}\frac{(n-1)(3n+1)\,\Gamma(2n+1)}{\Gamma(n+2)\Gamma(n+3)}\cdot B\left(n+\frac{1}{2},n+\frac{1}{2}\right)\\&=&\frac{4}{\pi}\int_{0}^{1}f\left(\frac{u(1-u)}{1-u+u^2}\right)\,\frac{du}{\sqrt{u(1-u)}}\tag{3}\end{eqnarray*}$$
where:
$$ f(x)=\sum_{n\geq 1}\frac{(n-1)(3n+1)\,\Gamma(2n+1)}{\Gamma(n+2)\Gamma(n+3)}\,x^n=\frac{5-5 \sqrt{1-4 x}+2 x^2-2 x \left(4+\sqrt{1-4 x}+8\log 2\right)+16 x\,\log\left(1+\sqrt{1-4 x}\right)}{4 x^2}\tag{4}$$
gives, after some extra work, 
$$ \sum_{n\geq 1}\frac{C_{n+1}-2C_n}{4^n}\sum_{k\geq n}\frac{C_{k+1}-C_k}{4^k} = \color{red}{\frac{592}{3\pi}-62}.\tag{5}$$
