Gauss error function of square root function Why does the following equation hold?  
$$\sum_{i=1}^{\infty}\frac{2^i}{\sqrt{\pi}\prod_{j=1}^{i}(2j-1)}x^{\frac{2i-1}{2}}=e^x Erf(\sqrt{x})$$ 
in which $Erf$ is the Gauss error function.

\begin{align}
e^{x^2}\operatorname{erf}(x)
&=\frac{1}{\sqrt{\pi}}\int_{-x}^{x}e^{x^2-t^2}\,dt
&&\text{[definition of $\operatorname{erf}(x)$]}
\\&=\frac{2x}{\sqrt{\pi}}\int_{0}^{1}e^{4x^2y(1-y)}\,dy
&&\text{[substitution $t=x(1-2y)$]}
\\&=\frac{2x}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(4x^2)^n}{n!}\int_{0}^{1}y^n(1-y)^n\,dy
&&\text{[power series expansion of $e^z$]}
\\&=\frac{2x}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(4x^2)^n}{n!}\frac{n!^2}{(2n+1)!}
&&\text{[known "beta" integral]}
\\&=\frac{2x}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(2x^2)^n}{(2n+1)!!}=\frac{1}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{2^{n+1}x^{2n+1}}{(2n+1)!!}.
&&\text{[cancellation]}
\end{align}
