Challenge: Prove $\displaystyle \sum_{n \in \mathbb{N}} \frac{n!}{(2n+1)!}=e^{1/4}\sqrt{\pi}\ \ \mathrm{erf}(\frac{1}{2})$ I stumbled upon this cute sum while messing about, and I want to see what other solutions people propose before I put forward my own (which may be unnecessarily complicated).
You can use any maths you like.
NB: $\displaystyle \mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$
 A: Recall for all $x > 0$,
$$\sinh x = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}
\quad\implies\quad
\frac{\sinh \sqrt{x}}{\sqrt{x}} = \sum_{n=0}^\infty \frac{x^n}{(2n+1)!}
$$
and the integral representation of $n!$
$$n! = \int_0^\infty x^n e^{-x}dx$$
The sum at hand can be rewritten as
$$
\sum_{n=0}^\infty \int_0^\infty\frac{x^{n}}{(2n+1)!}  e^{-x}dx
= \int_0^\infty e^{-x}\left(\sum_{n=0}^\infty \frac{x^n}{(2n+1)!}\right)dx
= \int_0^\infty e^{-x}\frac{\sinh\sqrt{x}}{\sqrt{x}}dx
$$
Change variable to $t = \sqrt{x}$, this reduces to
$$\int_0^\infty e^{-t^2}(e^t - e^{-t}) dt
= e^{\frac14}\int_0^\infty \left(e^{-(t-\frac12)^2} - e^{-(t+\frac12)^2}\right)dt
= e^{\frac14}\left(\int_{-\frac12}^\infty - \int_{\frac12}^\infty\right) e^{-t^2}dt\\
= e^{\frac14}\int_{-\frac12}^{\frac12} e^{-t^2}dt
= 2e^{\frac14}\int_0^{\frac12} e^{-t^2}dt
= e^{\frac14}\sqrt{\pi}\,\mathrm{erf}\left(\frac12\right)
$$
A: Simply, by using Euler's beta function,
$$S=\sum_{n\geq 0}\frac{n!}{(2n+1)!}=\sum_{n\geq 0}\frac{B(n+1,n+1)}{n!}=\int_{0}^{1}\sum_{n\geq 0}\frac{x^n(1-x)^n}{n!}\,dx = \int_{0}^{1}e^{x(1-x)}\,dx$$
and by setting $x=t+\frac{1}{2}$
$$ S = \int_{0}^{1}e^{x(1-x)}\,dx = 2e^{1/4} \int_{0}^{1/2}e^{-t^2}\,dt.$$
A: Outline:


*

*We know that $$e^{1/4} = \sum_{n=0}^\infty \frac{1}{2^{2n} n!}$$

*and that $$\begin{align}
\sqrt{\pi}\operatorname{erf}\left(\frac{1}{2}\right) &= \int_0^{1/2} e^{-t^2}dt 
= \int_0^{1/2} \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!} dt
= \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\int_0^{1/2} t^{2n} dt
\\
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{n! 2^{2n}} \frac{1}{2n+1}
\end{align}$$
so that, writing it as a product of series and using the Cauchy product of these,
$$\begin{align}
\sqrt{\pi}\operatorname{erf}\left(\frac{1}{2}\right) e^{1/4} 
&= \sum_{n=0}^{\infty} \frac{(-1)^n}{n! 2^{2n}} \frac{1}{2n+1}\sum_{n=0}^\infty \frac{1}{2^{2n} n!}
= \sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^k}{k! 2^{2k} (2k+1)}\frac{1}{(n-k)!2^{2n-2k}} \\
&= \sum_{n=0}^\infty \frac{1}{2^{2n}n!}\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{k! (2k+1)}
= \sum_{n=0}^\infty \frac{1}{2^{2n}n!}\frac{(2n)!!}{(2n+1)!!}
\end{align}$$
and from there it is not hard to conclude from the properties of the double factorial.
(If needed, I can fill in the remaining steps later.)

Finishing it:
$$\begin{align}
\frac{(2n)!!}{(2n+1)!!}
= \frac{2^n n!}{(2n+1)!!}
= \frac{2^n n!\sqrt{\pi}}{2^{n+1}\Gamma(\frac{1}{2}+n+1)}
\end{align}$$
from which, using the properties of the Gamma function,
$$\begin{align}
\frac{(2n)!!}{2^{2n}n!(2n+1)!!}
&= \frac{\sqrt{\pi}}{2^{2n+1}\Gamma(\frac{1}{2}+n+1)}
= \frac{\sqrt{\pi}2^{2n+2}(n+1)!}{2^{2n+1}\sqrt{\pi}(2n+2)!}\\
&= \frac{2 (n+1)n!}{2(n+1)(2n+1)!}
= \frac{n!}{(2n+1)!}
\end{align}$$
and we are done.
A: Let $\displaystyle I_n= \int_0^{\infty} x^ne^{-x^2}dx$.
Let $S=\displaystyle \sum_{n \geq 0} \frac{I_n}{n!}=\int_0^{\infty}e^xe^{-x^2}dx=e^{1/4}\int_0^{\infty} e^{-(x-1/2)^2}dx=\frac{1}{2}e^{1/4}\sqrt{\pi}(\mathrm{erf}(\frac{1}{2})+1)$.
Splitting $S$ into odd and even parts, have
$\displaystyle S_{\mathrm{even}}=\sum_{n \geq 0} \int_0^{\infty} \frac{x^{2n}}{2n!}e^{-x^2}dx=\frac{1}{2}\int_0^{\infty} (e^{x}+e^{-x})e^{-x^2}dx=\frac{1}{2}e^{1/4}\sqrt{\pi}$
So $\displaystyle S_{\mathrm{odd}}=\frac{1}{2} e^{1/4}\sqrt{\pi}\ \ \mathrm{erf}(\frac{1}{2})$.
Conclude using $\displaystyle I_{2n+1}=\frac{n!}{2}$ (from integration by parts). 
We get $\displaystyle \sum_{n \geq 0} \frac{n!}{(2n+1)!}=e^{1/4}\sqrt{\pi}\ \ \mathrm{erf}(\frac{1}{2})$.
