Closed form of a power series involving double factorial: $\sum_{n=0}^{\infty} \frac{x^n}{(2n-1)!!}$ I meet a series of the form
$$\sum_{n=0}^{\infty} \frac{x^n}{(2n-1)!!}$$
where $(-1)!! = 1$.
I guess it is a Taylor expansion of a function but I don't know what it is. Could anyone here help me?
Remark: The problem comes from calculating a renewal process. Assume $N(t)$ is a renewal process with interarrival time $X_i$ where $X_i$ i.i.d. follow $\chi^2_1$. Then the arrival time of the $k$th event is $S_k \sim \chi^2_k$. Then the renewal function is
$$m(t) = \mathbb{E}N(t) =\sum_{k=1}^\infty Pr(S_k \leq t)$$
which is
$$\sum_{k=1}^\infty \int_0^t \frac{x^{k/2-1}e^{-x/2}}{2^{k/2}\Gamma(k/2)}dx.$$
We can exchange the summation and the integral and divide the summation into two parts according to $k$ is even or odd.
The part for $k$ is even is easy. But for $k$ is odd, I think we need to deal the series in the beginning of the problem.
 A: Your series is certainly not a trivial one. We shall prove that your series is given by $1+f(\sqrt{x})$, where
\begin{equation}
f(x)=xe^{x^2/2}\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right)
\end{equation}
and $\text{erf}(z)$ is the error function. We will be making use of the Taylor series for $\text{erf}(z)$ which is as follows:
\begin{equation}
\text{erf}(z)=\frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{(-1)^nz^{2n+1}}{n!(2n+1)}
\end{equation}
Now, we expand Taylor series of both $e^{x^2/2}$ and $\text{erf}\left(\frac{x}{\sqrt{2}}\right)$, and take the Cauchy product:
\begin{equation}
\begin{split}
f(x)&=xe^{x^2/2}\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right)\\
&=x\sqrt{\frac{\pi}{2}}\left[\sum_{k=0}^\infty \frac{x^{2k}}{2^kk!}\right]\left[\sqrt{\frac{2}{\pi}}\sum_{\ell=0}^\infty \frac{(-1)^kx^{2\ell+1}}{2^\ell \ell ! (2\ell+1)}\right]\\
&=x^2\left[\sum_{k=0}^\infty \frac{x^{2k}}{2^kk!}\right]\left[\sum_{\ell=0}^\infty \frac{(-1)^kx^{2\ell}}{2^\ell \ell ! (2\ell+1)}\right]\\
&=x^2\sum_{n=0}^\infty\sum_{k=0}^n\frac{x^{2n-2k}}{2^{n-k}(n-k)!}\cdot\frac{(-1)^kx^{2k}}{2^kk!(2k+1)}\\
&=x^2\sum_{n=0}^\infty\frac{x^{2n}}{2^nn!}\sum_{k=0}^n{n\choose k}\frac{(-1)^k}{2k+1}\\
\end{split}
\end{equation}
We can evaluate this last sub-sum using the Beta function:
\begin{equation}
\begin{split}
\sum_{k=0}^n{n\choose k}\frac{(-1)^k}{2k+1}&=\sum_{k=0}^n{n\choose k}(-1)^k\int_0^1t^{2k}dt\\
&=\int_0^1\sum_{k=0}^n{n\choose k}(-1)^kt^{2k}dt\\
&=\int_0^1(1-t^2)^ndt\\
&=\int_0^1\frac{(1-u)^n}{2u^{1/2}}du\\
&=\frac{1}{2}B(1/2,n+1)\\
&=\frac{\Gamma(1/2)\Gamma(n+1)}{2\Gamma(n+3/2)}\\
&=\frac{2^nn!}{(2n+1)!!}
\end{split}
\end{equation}
We may therefore simplify
\begin{equation}
f(x)=x^2\sum_{n=0}^\infty\frac{x^{2n}}{(2n+1)!!}=\sum_{n=1}^\infty\frac{x^{2n}}{(2n-1)!!}
\end{equation}
which means that
\begin{equation}
\sum_{n=0}^\infty\frac{x^n}{(2n-1)!!}=1+f(\sqrt{x})=1+e^{x/2}\sqrt{\frac{\pi x}{2}}\text{erf}\left(\sqrt{\frac{x}{2}}\right)
\end{equation}
as desired.
A: To compute
$$
f(x)=\sum_{n=0}^\infty\frac{x^n}{(2n-1)!!}\tag1
$$
consider
$$
\begin{align}
g(x)
&=\frac{f(2x)}{\sqrt{2x}}\tag{2a}\\
&=\sum_{n=0}^\infty\frac{(2x)^{n-1/2}}{(2n-1)!!}\tag{2b}
\end{align}
$$
where
$$
\begin{align}
g'(x)
&=\sum_{n=0}^\infty\frac{(2x)^{n-3/2}}{(2n-3)!!}\tag{3a}\\
&=g(x)-(2x)^{-3/2}\tag{3b}
\end{align}
$$
since $(-3)!!=-1$.
Applying an integrating factor of $e^{-x}$ to $(3)$ yields
$$
(e^{-x}g(x))'=-e^{-x}(2x)^{-3/2}\tag4
$$
and thus,
$$
\begin{align}\newcommand{\erf}{\operatorname{erf}}
g(x)
&=\frac{e^x}{2\sqrt2}\int_x^\infty e^{-t}t^{-3/2}\,\mathrm{d}t+ce^x\tag{5a}\\
&=-\frac{e^x}{\sqrt2}\int_x^\infty e^{-t}\,\mathrm{d}t^{-1/2}+ce^x\tag{5b}\\
&=\frac1{\sqrt{2x}}-\frac{e^x}{\sqrt2}\int_x^\infty e^{-t}t^{-1/2}\,\mathrm{d}t+ce^x\tag{5c}\\
&=\frac1{\sqrt{2x}}-\sqrt2e^x\int_{\sqrt{x}}^\infty e^{-t^2}\,\mathrm{d}t+ce^x\tag{5d}\\
&=\frac1{\sqrt{2x}}+\sqrt2e^x\int_0^{\sqrt{x}}e^{-t^2}\,\mathrm{d}t\tag{5e}\\
&=\frac1{\sqrt{2x}}+e^x\sqrt{\frac\pi2}\erf\left(\sqrt{x}\right)\tag{5f}
\end{align}
$$
Explanation:
$\text{(5a):}$ solve $(4)$ for $g(x)$ where $c$ is a constant TBD
$\text{(5b):}$ prepare to integrate by parts
$\text{(5c):}$ integrate by parts
$\text{(5d):}$ substitute $t\mapsto t^2$
$\text{(5e):}$ set $c=\sqrt{\pi/2}=\sqrt2\int_0^\infty e^{-t^2}\mathrm{d}t$
$\phantom{\text{(5e):}}$ so that $g$ is an odd function of $\sqrt{x}$
$\text{(5f):}$ $\int_0^xe^{-t^2}\,\mathrm{d}t=\frac{\sqrt\pi}2\erf(x)$
Reversing $\text{(2a)}$ by setting $f(x)=\sqrt{x}\,g(x/2)$, we get
$$
f(x)=1+e^{x/2}\sqrt{\frac{\pi x}2}\erf\left(\sqrt{x/2}\right)\tag6
$$
A: Using
$$(2 n -1)!! = \frac{(2 n)!}{2^n \, n!} = 2^n \, \left(\frac{1}{2}\right)_{n}$$
then
$$ \sum_{n=0}^{\infty} \frac{x^n}{(2 n - 1)!!} = \sum_{n=0}^{\infty} \frac{\left(\frac{x}{2}\right)^n}{(1/2)_{n}} = {}_{1}F_{1}\left(1 ; \frac{1}{2} ; \frac{x}{2} \right),$$
where $(a)_{n}$ is the Pochhammer symbol, ${}_{1}F_{1}(a; b; x)$ is the confluent hypergeometric function. Now, by using
$$ {}_{1}F_{1}\left(1 ; \frac{1}{2} ; x \right) = 1 + \sqrt{\pi \, x} \, e^{x} \, \text{erf}(\sqrt{x}) $$
then
$$ \sum_{n=0}^{\infty} \frac{x^n}{(2 n - 1)!!} = 1 + \sqrt{\frac{\pi \, x}{2}} \, e^{x/2} \, \text{erf}\left(\sqrt{\frac{x}{2}}\right). $$
