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

Just as the topic ask how to evaluate $$\sum_{k=1}^{\infty}\frac{(18)[(k-1)!]^2}{(2k)!}.$$

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–  Pedro Tamaroff Jun 18 '12 at 3:37

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

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 $$f\left(\frac{1}{4} \right) = \int_0^{\frac{1}{4}} 2 \frac{\mathrm{d}}{\mathrm{d} t} \arcsin^2(\sqrt{t}) = 2 \arcsin^2\left(\frac{1}{2}\right) = \frac{\pi^2}{18}$$

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That last step is still perhaps a little miraculous, but this looks like the way to go... –  Steven Stadnicki Jan 12 '12 at 18:26
Would you please elaborate on how you got that last step? I had to approach this in a wholly different manner. –  robjohn Apr 6 '12 at 12:48
@robjohn I have filled in the derivation. It might be not that different from yours. +1 on yours, by the way. –  Sasha Apr 6 '12 at 15:10
@Sasha: that does look similar mod a change of variables. We also dropped to the derivative at different points. (+1) –  robjohn Apr 6 '12 at 15:41
Maple gives the answer as $36 \arcsin(1/2)^2$. More generally, $$\sum_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} t^k = 2 \arcsin \left(\frac{\sqrt{t}}{2}\right)^2$$
@Robert Israel: I guess there is a t^k missing on the left hand side and $36 \arcsin(1/2)^2 = \pi^2$ –  Fabian Jan 12 '12 at 17:56
Yes, start with Robert's power series $$F(x) = \sum_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} x^k,$$ fiddle with it to verify that it satisfies differential equation $$\bigl(4 x - x^{2}\bigr)\frac{d^{2} F (x)}{d x^{2}} + (2 - x) \frac{d F (x)}{d x} - 1 = 0$$ and solve that DE. –  GEdgar Jan 12 '12 at 18:38