Prove $\sum_{n=1}^{\infty} \frac{n!}{(2n)!} = \frac{1}{2}e^{1/4} \sqrt{\pi} \text{erf}(\frac{1}{2})$

I would like to prove:

$$\sum_{n=1}^{\infty} \frac{n!}{(2n)!} = \frac{1}{2}e^{1/4} \sqrt{\pi} \text{erf}(\frac{1}{2})$$

What I did was consider:

$$e^{-t^2}=\sum_{n=0}^{\infty} (-1)^n \frac{t^{2n}}{n!}$$

Then integrate term by term from $0$ to $x$ to get:

$$\frac{\sqrt{\pi}}{2}\text{erf} (x)=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(2n+1)}$$

Then I substituted in $x=\frac{1}{2}$ and tried some manipulations but didn’t get anywhere. May someone help, thanks.

Solutions Collecting From Web of "Prove $\sum_{n=1}^{\infty} \frac{n!}{(2n)!} = \frac{1}{2}e^{1/4} \sqrt{\pi} \text{erf}(\frac{1}{2})$"

We have $\frac{n!}{(2n)!}=\frac{B(n,n+1)}{(n-1)!}$, hence:

$$\sum_{n\geq 1}\frac{n!}{(2n)!} = \int_{0}^{1}\sum_{n\geq 1}\frac{x^{n-1}(1-x)^n}{(n-1)!}\,dx = \int_{0}^{1}(1-x)e^{x(1-x)}\,dx$$
and the result follows by setting $x=t+\frac{1}{2}$ in the last integral, leading to $e^{1/4}\int_{0}^{\frac{1}{2}}e^{-u^2}\,du.$