Closed form of a power series involving double factorial: $\sum_{n=0}^{\infty} \frac{x^n}{(2n-1)!!}$

Question

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.

Answer 1

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)^\ell x^{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)^\ell x^{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.

Answer 2

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 $$ \sum_{n=0}^{\infty} \frac{x^n}{(2 n - 1)!!} = 1 + \sqrt{\frac{\pi \, x}{2}} \, e^{x/2} \, \text{erf}\left(\sqrt{\frac{x}{2}}\right). $$

Answer 3

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 $$ f(x)=1+e^{x/2}\sqrt{\frac{\pi x}2}\erf\left(\sqrt{x/2}\right)\tag6 $$