Cómo evaluar esta doble suma infinita (catalán número)

Question

Deje $C_n = \dfrac{1}{n+1}\binom{2n}{n}$.

Es posible encontrar el valor exacto de esta suma infinita ? $$\sum_{n=1}^\infty \sum_{k=n}^\infty \frac{\left(C_{n+1}-2C_n\right)\left(C_{k+1}-C_k\right)}{4^{n+k}}$$

En general, es posible evaluar $$\sum_{n=1}^\infty \sum_{k=n}^\infty \frac{\left(C_{n+1}-2C_n\right)\left(C_{k+1}-C_k\right)}{x^{n+k}}$$ en el plazo de $x$?

Gracias en avance.

Answer 1

Tenemos: $$ C_{k+1}-C_k = \frac{3k\, 4^k\, \Gamma\left(k+\frac{1}{2}\right)}{\sqrt{\pi}\,\Gamma(k+3)}\tag{1} $$ por lo tanto: $$ \sum_{k\geq n}\frac{C_{k+1}-C_k}{4^k} =\frac{(6n+2)\,\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}\,\Gamma(n+2)} \tag{2}$$ y: $$\begin{eqnarray*} \sum_{n\geq 1}\frac{C_{n+1}-2C_n}{4^n}\sum_{k\geq n}\frac{C_{k+1}-C_k}{4^k} &=& \frac{4}{\pi}\sum_{n\geq 1}\frac{(n-1)\,\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+3)}\cdot \frac{(3n+1)\,\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+2)}\\&=&\frac{4}{\pi}\sum_{n\geq 1}\frac{(n-1)(3n+1)\,\Gamma(2n+1)}{\Gamma(n+2)\Gamma(n+3)}\cdot B\left(n+\frac{1}{2},n+\frac{1}{2}\right)\\&=&\frac{4}{\pi}\int_{0}^{1}f\left(\frac{u(1-u)}{1-u+u^2}\right)\,\frac{du}{\sqrt{u(1-u)}}\tag{3}\end{eqnarray*}$$ donde: $$ f(x)=\sum_{n\geq 1}\frac{(n-1)(3n+1)\,\Gamma(2n+1)}{\Gamma(n+2)\Gamma(n+3)}\,x^n=\frac{5-5 \sqrt{1-4 x}+2 x^2-2 x \left(4+\sqrt{1-4 x}+8\log 2\right)+16 x\,\log\left(1+\sqrt{1-4 x}\right)}{4 x^2}\tag{4}$$ da, después de algo más de trabajo,

$$ \sum_{n\geq 1}\frac{C_{n+1}-2C_n}{4^n}\sum_{k\geq n}\frac{C_{k+1}-C_k}{4^k} = \color{red}{\frac{592}{3\pi}-62}.\tag{5}$$