Sum: $\sum\limits_{n=0}^\infty \frac{n!}{(2n)!}$

Here's a hint. Write the fraction as $\int_0^\infty\frac{x^n}{(2n)!}e^{-x}dx$. The sum is then $\int_0^\infty\cosh\sqrt{x}e^{-x} \, dx=\int_0^\infty 2y\cosh y e^{-y^2} \, dy$. Can you take it from there?

One might note that:

$$\frac{n!}{(2n)!}=\frac{\sqrt\pi}{4^n\Gamma(n+1/2)}$$

Indeed, this makes your series a special case of the following function:

$$f_\alpha(x)=\sum_{n=0}^\infty\frac{x^n}{\Gamma(n+\alpha)}$$

With $S=\sqrt\pi f_{1/2}(1/4)$.

Wonderfully, this problem serves as a good introduction for fractional derivatives. Let us define here the following:

$$\frac{\mathrm d^\alpha}{\mathrm dx^\alpha}f(x)=\frac1{\Gamma(1-\{\alpha\})}\frac{\mathrm d^{\lceil\alpha\rceil}}{\mathrm dx^{\lceil\alpha\rceil}}\int_0^x(x-t)^{-\{\alpha\}}f(t)~\mathrm dt$$

where $\lceil\alpha\rceil$ is the ceiling function and $\{\alpha\}=\alpha-\lfloor\alpha\rfloor$ is the fractional part of $\alpha$. From the above definition, one can deduce that

$$\frac{\mathrm d^\alpha}{\mathrm dx^\alpha}x^\beta=\frac{x^{\beta-\alpha}\Gamma(\beta+1)}{\Gamma(\beta-\alpha+1)}$$

From this, one may deduce that

$$\frac{x^n}{\Gamma(n+\alpha)}=x^{1-\alpha}\frac{\mathrm d^{1-\alpha}}{\mathrm dx^{1-\alpha}}\frac{x^n}{n!}$$

Thus, we find that

$$f_\alpha(x)=x^{1-\alpha}\frac{\mathrm d^{1-\alpha}}{\mathrm dx^{1-\alpha}}e^x$$

In the particular case of $\alpha=1/2$,

$$\frac{\mathrm d^{1/2}}{\mathrm dx^{1/2}}e^x=\frac1{\Gamma(1/2)}\int_0^x(x-t)^{-1/2}e^t~\mathrm dt$$

Let $x-t=u^2$ to get

$$\frac{\mathrm d^{1/2}}{\mathrm dx^{1/2}}e^x=\frac1{\Gamma(3/2)}\int_0^{\sqrt x}u^2e^{x-u^2}~\mathrm du$$

Whereupon the relationship to the error function is immediate.

The general case for $f_\alpha(x)$ is left as extra credit for the reader with the additional hint that incomplete Gamma functions are recommended.

The series can be written also in terms of the Incomplete Gamma function. As noted by Simply Beautiful Art we have $$\sum_{n\geq0}\frac{n!}{\left(2n\right)!}=\sum_{n\geq0}\frac{\Gamma\left(1/2\right)}{4^{n}\Gamma\left(n+1/2\right)}$$ $$=1+\frac{1}{4}\sum_{n\geq0}\frac{\Gamma\left(1/2\right)}{4^{n}\Gamma\left(n+1+1/2\right)}=1+\frac{1}{4}\sum_{n\geq0}\frac{1}{4^{n}\left(1/2\right)_{n+1}}$$ where $\left(x\right)_{n}=\Gamma\left(x\right)/\Gamma\left(x+n\right)$ is the Pochhammer symbol and since $$\gamma\left(a,z\right)e^{z}z^{-a}=\sum_{n\geq0}\frac{z^{n}}{\left(a\right)_{n+1}},\,a\neq-k,\,k\in\mathbb{N}$$ we have $$\sum_{n\geq0}\frac{n!}{\left(2n\right)!}=\color{red}{1+\frac{e^{1/4}\gamma\left(1/2,1/4\right)}{2}}\approx \color{blue}{1.5923}$$