Evaluar $\sum\limits_{k=1}^{\infty}\frac{(18)[(k-1)!]^2}{(2k)!}$

Question

Así como el tema de la pregunta de cómo evaluar $$\sum_{k=1}^{\infty}\frac{(18)[(k-1)!]^2}{(2k)!}.$$

Answer 1

Observe que $$ \frac{(k-1)!^2}{(2k)!} = \frac{\Gamma(k) \Gamma(k)}{ \Gamma(2k+1) } = \frac{1}{2k} B(k,k) = \frac{1}{2k} \int_0^1 x^{k-1} (1-x)^{k-1} \mathrm{d} x $$ Así $$ \begin{eqnarray} \sum_{k=1}^\infty \frac{(k-1)!^2}{(2k)!} &=& \int_0^1 \left( \sum_{k=1}^\infty \frac{1}{2k} x^{k-1} (1-x)^{k-1} \right) \mathrm{d} x = \int_0^1 \frac{-\ln(1-x+x^2)}{2x (1-x)} \mathrm{d} x \\ &=& -2 \int_{-1/2}^{1/2} \frac{\ln(3/4+u^2)}{1-4 u^2} \mathrm{d} u = -2 \int_0^1 \frac{\ln((3 +u^2)/4)}{1-u^2} \mathrm{d} u = \frac{\pi^2}{18} \end{eqnarray} $$

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Agregó
Con el fin de llenar en la última igualdad, definir $$ f(t) = -2 \int_0^1 \frac{\ln\left(1 - t(1-u^2) \right)}{1-u^2} \mathrm{d} u $$ Claramente $f(0) = 0$, y estamos interesados en calcular los $f\left(\frac{1}{4} \right)$. $$ f^\prime(t) = 2 \int_0^1 \frac{\mathrm{d} u}{1 + t(1-u^2)} \stackrel{{u = \sqrt{\frac{1-t}{t}} \tan(\phi)}}{=} \int_0^{\arcsin(\sqrt{t})} \frac{2 \mathrm{d} \phi}{\sqrt{t(1-t)}} = \frac{2 \arcsin(\sqrt{t})}{\sqrt{t(1-t)}} = \\ 2 \frac{\mathrm{d}}{\mathrm{d} t} \arcsin^2(\sqrt{t}) $$ Así $$ f\left(\frac{1}{4} \right) = \int_0^{\frac{1}{4}} 2 \frac{\mathrm{d}}{\mathrm{d} t} \arcsin^2(\sqrt{t}) = 2 \arcsin^2\left(\frac{1}{2}\right) = \frac{\pi^2}{18} $$

Answer 2

Arce da la respuesta como $36 \arcsin(1/2)^2$. De manera más general, $$ \sum_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} t^k = 2 \arcsin \left(\frac{\sqrt{t}}{2}\right)^2$$