Can we simplify this summation involving factorials? The sum is as follows:
$$\sum_{k=1}^n\frac{1}{(k!)^2(2n-2k)!}\frac{1}{2^{2k}}$$
Can we compute this to an expression without summation?
 A: You can hide the summation using
$$\sum_{k=1}^n\frac{{x^{2k}}}{(k!)^2(2n-2k)!}=\frac{\, _2F_1\left(\frac{1}{2}-n,-n;1;4 x^2\right)-1}{(2 n)!}$$ where appears the gaussian hypergeometric function.
S0, for $x=\frac 12$, you will get $$\frac{\, _2F_1\left(\frac{1}{2}-n,-n;1;1\right)-1}{(2 n)!}$$
More interesting is @Mark Viola's hint which leads to 
$$\frac{\frac{2^{2 n}\, \Gamma \left(2n+\frac{1}{2} \right)}{\sqrt{\pi }\, \Gamma (2
   n+1)}-1}{(2 n)!}$$
A: For any formal Laurent series in any indeterminate $z$, let $[z^k](\cdots)$ be a shorthand for the coefficient in front of $z^k$. More precisely,
$$[z^k] \sum_{i=-\infty}^\infty c_i z^i \stackrel{def}{=} c_k$$
The sum at hand can be rewritten as
$$\mathcal{S} \stackrel{def}{=} \sum_{k=1}^n\frac{1}{(k!)^2(2n-2k)!}\frac{1}{2^{2k}}
= \frac{1}{(2n)!}\sum_{k=1}^n \binom{2n}{2k}\binom{2k}{k}\frac{1}{2^{2k}}
$$
Recall $\displaystyle\;\binom{2k}{k} = [z^0] (z + z^{-1})^{2k}\;$ and $\;[z^0] (z+z^{-1})^\ell = 0\;$ for odd $\ell$, we have
$$\begin{align}
(2n)!\mathcal{S} &= [z^0] \sum_{k=1}^n \binom{2n}{2k}
\left(\frac{z + z^{-1}}{2}\right)^{2k}
= [z^0]\sum_{\ell=1}^{2n} \binom{2n}{\ell}\left(\frac{z+z^{-1}}{2}\right)^\ell\\
&= [z^0]\left\{\left(1 + \frac{z+z^{-1}}{2}\right)^{2n} - 1\right\}
\end{align}$$
Change variable to $z = \omega^2$, we obtain
$$
\mathcal{S} = \frac{1}{(2n)!}[\omega^0]\left\{\frac1{4^n}(\omega + \omega^{-1})^{4n} - 1\right\}
= \frac{1}{(2n)!}\left\{\frac{1}{4^n}\binom{4n}{2n} - 1\right\}
$$
