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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.]

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  • $\begingroup$ This series does not sum to $\exp x^2$, but to $1 + (1+\mathrm{erf} x) \exp x^2$, as we can see using WolframAlpha (wolframalpha.com/input/?i=sum+x%5Ei%2F(i%2F2)!) $\endgroup$ – rafa11111 Dec 18 '17 at 23:05
  • $\begingroup$ Furthermore, the first equality is wrong. The series is equal to $\sum_j x^{2j}/j! + \sum_j x^{2j+1}/(j+1/2)!$. However, both equalities holds if the first summation is evaluated only over even integers, naturally. $\endgroup$ – rafa11111 Dec 18 '17 at 23:10
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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, $$ \bbox[5px,border:2px solid #C0A000]{\sum_{n=0}^\infty\frac{x^n}{(n/2)!}=e^{x^2}(1+\operatorname{erf}(x))} $$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}:

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

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