Evaluate $\sum\limits_{k=1}^{\infty}\frac{(18)[(k-1)!]^2}{(2k)!}$ Just as the topic ask how to evaluate $$\sum_{k=1}^{\infty}\frac{(18)[(k-1)!]^2}{(2k)!}.$$
 A: Maple gives the answer as $36 \arcsin(1/2)^2$.  More generally,
$$ \sum_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} t^k = 2 \arcsin \left(\frac{\sqrt{t}}{2}\right)^2$$
A: Notice that 
$$
  \frac{(k-1)!^2}{(2k)!} = \frac{\Gamma(k) \Gamma(k)}{ \Gamma(2k+1) } = \frac{1}{2k} B(k,k) = \frac{1}{2k} \int_0^1 x^{k-1} (1-x)^{k-1} \mathrm{d} x
$$
Thus
$$ \begin{eqnarray}
   \sum_{k=1}^\infty \frac{(k-1)!^2}{(2k)!} &=& \int_0^1 \left( \sum_{k=1}^\infty \frac{1}{2k} x^{k-1} (1-x)^{k-1} \right) \mathrm{d} x = \int_0^1 \frac{-\ln(1-x+x^2)}{2x (1-x)} \mathrm{d} x  \\
   &=& -2 \int_{-1/2}^{1/2} \frac{\ln(3/4+u^2)}{1-4 u^2} \mathrm{d} u = -2 \int_0^1 \frac{\ln((3 +u^2)/4)}{1-u^2} \mathrm{d} u = \frac{\pi^2}{18}
 \end{eqnarray}
$$

Added
In order to fill in on the last equality, define
$$
 f(t) = -2 \int_0^1 \frac{\ln\left(1 - t(1-u^2) \right)}{1-u^2} \mathrm{d} u
$$
Clearly $f(0) = 0$, and we are interested in computing $f\left(\frac{1}{4} \right)$. 
$$
   f^\prime(t) = 2 \int_0^1 \frac{\mathrm{d} u}{1 + t(1-u^2)} \stackrel{{u = \sqrt{\frac{1-t}{t}} \tan(\phi)}}{=} \int_0^{\arcsin(\sqrt{t})} \frac{2 \mathrm{d} \phi}{\sqrt{t(1-t)}} =  \frac{2 \arcsin(\sqrt{t})}{\sqrt{t(1-t)}} = \\ 2  \frac{\mathrm{d}}{\mathrm{d} t} \arcsin^2(\sqrt{t})
$$
Thus
$$
   f\left(\frac{1}{4} \right) = \int_0^{\frac{1}{4}} 2  \frac{\mathrm{d}}{\mathrm{d} t} \arcsin^2(\sqrt{t}) = 2 \arcsin^2\left(\frac{1}{2}\right) = \frac{\pi^2}{18}
$$
