TITLE: Compute the series : $\sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!}$
QUESTION [13 upvotes]: How would you compute the following series? I'm interested in some easy approaches that would allow me to work it out.
$$\sum_{n=1}^{\infty} \frac{4^n n!}{(2n)!}$$

REPLY [12 votes]: Here is a generalization. Consider the generating function
$$f(x) = \sum_{n \ge 1} \frac{x^{2n} n!}{(2n)!}.$$
It's an easy check that $2$ is in the radius of convergence, so that we need to compute $f(2)$ to solve your problem. However, it's easy to see that
$$f''(x) = 1 + \sum_{n \ge 1} \frac{x^{2n} (n + 1)!}{(2n)!}$$
Moreover,
$$xf'(x) = 2 \sum_{n \ge 1} \frac{x^{2n} n!}{(2n)!} \cdot n$$
We get the second-order ODE
$$\frac{xf'(x)}{2} + f(x) = f''(x) - 1, f(0) = 0, f'(0) = 0.$$
And the solution
$$f(x) = \frac{1}{2} \cdot e^{x^2/4} \cdot x \sqrt{\pi} \mathrm{erf}(x/2).$$
Just plug in $x = 2$ to get your answer, $e\sqrt{\pi} \mathrm{erf}(1)$.