<s>[INST] 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: $\displaystyle \mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$
 [/INST] 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.
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