Infinite series with fraction factorial An infinite sum is given and reduced as follows:
$$ \sum_{j}\frac{1}{(j/2)!}x^{j}=\sum_{j}\frac{1}{j!}x^{2j}=e^{x^{2}}$$
The second step is clear, but I am not sure about the first step in this reduction. Namely, if we assume that $k=j/2$, to simplify the factorial in the denominator, the form will then look indeed like $\sum_{k}\frac{1}{k!}x^{2k}$, but the counter $k$ here must be multiples of halves ($k=1/2, 1, 3/2, 2, 5/3, \cdots$), instead of integers as in $j$ originally ($1, 2, 3, \cdots$). How can we treat this as the usual exponential series, which is based by definition on integers, to get the final answer as $e^{x^{2}}$?
[Update: forgot to say that this result is supposed to hold for large $x$ values.]
 A: For reference,
$$
\begin{align}
\left(n-\tfrac12\right)!
&=\left(-\tfrac12\right)!\frac{1\cdot3\cdots(2n-1)}{2^n}\\
&=\sqrt\pi\frac{(2n)!}{4^nn!}
\end{align}
$$

Let
$$
f(x)=\sum_{n=1}^\infty\frac{x^{n-\frac12}}{\left(n-\frac12\right)!}
$$
Then
$$
f'(x)=f(x)+\frac{x^{-1/2}}{\sqrt\pi}
$$
Multiply by an integrating factor of $e^{-x}$
$$
(e^{-x}f(x))'=e^{-x}\frac{x^{-1/2}}{\sqrt\pi}
$$
Integrating from $0$, since $f(0)=0$, gives
$$
\begin{align}
e^{-x}f(x)
&=\frac1{\sqrt\pi}\int_0^x e^{-t}t^{-1/2}\,\mathrm{d}t\\
&=\frac2{\sqrt\pi}\int_0^{\sqrt x} e^{-u^2}\,\mathrm{d}u\\[6pt]
&=\operatorname{erf}\left(\sqrt x\right)
\end{align}
$$
Thus,
$$
\sum_{n=1}^\infty\frac{x^{n-\frac12}}{\left(n-\frac12\right)!}=e^x\operatorname{erf}\left(\sqrt x\right)
$$
Substituting $x\mapsto x^2$, the series above gives the terms with odd exponent, while $e^{-x^2}$ provides the terms with even exponent. Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{n=0}^\infty\frac{x^n}{(n/2)!}=e^{x^2}(1+\operatorname{erf}(x))}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{j = 0}^{\infty}{x^{j} \over \pars{j/2}!} & =
\sum_{j = 0}^{\infty}{\pars{x^2}^{j} \over j!} +
\sum_{j = 0}^{\infty}{x^{2j + 1} \over \pars{j + 1/2}!} = \expo{x^{2}} + 
\bbox[10px,#ffd]{\ds{\sum_{j = 0}^{\infty}
{x^{2j + 1} \over \Gamma\pars{j + 3/2}}}}\label{1}\tag{1}
\end{align}

\begin{align}
&\bbox[10px,#ffd]{\ds{\sum_{j = 0}^{\infty}
{x^{2j + 1} \over \Gamma\pars{j + 3/2}}}} =
\sum_{j = 0}^{\infty}
{x^{2j + 1} \over \Gamma\pars{\bracks{j + 1} + 1/2}}
\\[5mm] = &\
\sum_{j = 0}^{\infty}
{x^{2j + 1} \over \Gamma\pars{2j + 2}/
\bracks{\pars{2\pi}^{-1/2}\, 2^{2j + 3/2}\,\Gamma\pars{j + 1}}}\qquad
\pars{\begin{array}{l}
\ds{\Gamma}\mbox{-}\,Duplication\ Formula\ \mbox{in the}
\\
\mbox{denominator}.
\end{array}}
\\[5mm] = &\
{2 \over \root{\pi}}\,x\sum_{j = 0}^{\infty}{\Gamma\pars{j + 1} \over
\Gamma\pars{2j + 2}}\,\pars{4x^{2}}^{j} =
{2 \over \root{\pi}}\,x\sum_{j = 0}^{\infty}{\pars{4x^{2}}^{j} \over j!}\,
{\Gamma\pars{j + 1}\Gamma\pars{j + 1} \over\Gamma\pars{2j + 2}}
\\[5mm] = &\
{2 \over \root{\pi}}\,x\sum_{j = 0}^{\infty}{\pars{4x^{2}}^{j} \over j!}\,
\int_{0}^{1}t^{j}\pars{1 - t}^{j}\,\dd t\qquad
\pars{\begin{array}{l}
\mbox{The integral is the}
\\
Beta\ Function\ \mbox{value}\ \mrm{B}\pars{j + 1,j + 1}
\end{array}}
\\[5mm] = &\
{2 \over \root{\pi}}\,x\int_{0}^{1}\sum_{j = 0}^{\infty}
{\bracks{4x^{2}t\pars{1 - t}}^{j} \over j!}\,\dd t =
{2 \over \root{\pi}}\,x\int_{0}^{1}\exp\pars{-4x^{2}t\pars{t - 1}}\,\dd t
\\[5mm] = &\
{2 \over \root{\pi}}\,x\int_{0}^{1}\exp\pars{-4x^{2}
\bracks{\pars{t - {1 \over 2}}^{2} - {1 \over 4}}}\,\dd t =
{4\expo{x^{2}} \over \root{\pi}}\,x\,\int_{0}^{1/2}
\exp\pars{-4x^{2}t^{2}}\,\dd t
\\[5mm] = &\
\expo{x^{2}}\
\underbrace{{2 \over \root{\pi}}\int_{0}^{x}
\exp\pars{-t^{2}}\,\dd t}_{\ds{\mrm{erf}\pars{x}}} =
\bbx{\expo{x^{2}}\,\mrm{erf}\pars{x}}
\end{align}


With \eqref{1}:

$$
\bbx{\sum_{j = 0}^{\infty}{x^{j} \over \pars{j/2}!} =
\expo{x^{2}}\bracks{1 + \mrm{erf}\pars{x}}}
$$
