Closed Form involving double factorial Is there a closed form of
$$1+\frac{x}{3}+\frac{x^2}{5\cdot 3\cdot 1}+\frac{x^3}{7\cdot 5\cdot 3\cdot 1}+ \cdots = \sum_{n=0}^{\infty}\frac{x^n}{(2n+1)!!}$$
where $n!!$ is the double factorial?
Thank you very much!
According to WolframAlpha,
$\sum_{n=0}^{\infty}\frac{x^n}{(2n+1)!!}=\sqrt{\frac{\pi}{2x}}\,e^{x/2}\,\mathrm{erf}\left(\sqrt{\frac{x}{2}}\right)$.
Since 1, $1*3$, $1*3*5$  is the double factorial definition,
$\sum_{n=0}^{\infty}\frac{x^n}{(2n+1)!!}$ this is the right form....the equation you wrote is correct.
From wikipedia about the double factorial

and for odd n it is ${\displaystyle n!!=\prod _{k=1}^{\frac
&gt; {n+1}{2}}(2k-1)=n(n-2)(n-4)\cdots 3\cdot 1\,.}$

However, I hardly see how the double factorial form help you