This series is known? $\sum _{n=1}^{\infty }\frac{2^{n-1}x^{2n-1}}{\prod _{i=1}^n\left(2i-1\right)}$

Question

This series is known? I was solving a differential equation by power series and my solution involves that series

$$\sum _{n=1}^{\infty }\frac{2^{n-1}x^{2n-1}}{\prod _{i=1}^n\left(2i-1\right)}$$

Answer 1

My Wolfram Mathematica code

Sum[2^(n-1) x^(2 n-1)/Product[2i-1, {i,1,n}],
   {n,1,Infinity}] // InputForm

returns the result

(E^x^2*Sqrt[Pi]*Erf[x])/2

which is

$$ f(x) := \frac12 \sqrt{\pi } e^{x^2} \text{erf}(x) $$

and where $\,\text{erf}(x)\,$ is the Error function. The function $\,f(x)\,$ satisfies the ordinary differential equation

$$ f'(x) = 1 + 2\,xf(x). $$

By the way, this equation is the basis for an iterative recursion for the truncated power series of the function using

$$ f_0(x) := O(x), \quad f_{n+1}(x) = \int_0^x 1 + 2\,tf_n(t)\,dt. $$

The function also has another property. Define the differential operator

$$ T[ f(x) ] := e^{x^2} \frac{d}{dx} \left(e^{-x^2} f(x)\right) = f'(x) - 2\,x f(x). $$ Then your function satisfies $\, T[ f(x) ] = 1.$

Answer 2

Note that $$\begin{align} \frac1{\prod_{i=1}^n(2i-1)}&=\frac{\prod_{i=1}^n(2i)}{\left(\prod_{i=1}^n(2i)\right)\left(\prod_{i=1}^n(2i-1)\right)}\\ &=\frac{\prod_{i=1}^n(2i)}{\prod_{i=1}^n(2i(2i-1))}\\ &=\frac{2^nn!}{(2n)!}. \end{align}$$ So your sum is $$f(x)=\sum_{n\ge1}\frac{n!}{(2n)!}(2x)^{2n-1}=\sum_{n\ge0}\frac{2^{2n}n!}{(2n)!}\frac{x^{2n+1}}{2n+1}.$$ Let $t_n=\frac{2^{2n}n!}{(2n)!}\frac{x^{2n}}{2n+1}$, so that $f(x)=x\sum_{n\ge0}t_n$. Then $$\begin{align} \frac{t_{n+1}}{t_n}&=\frac{\frac{2^{2n+2}(n+1)!}{(2n+2)!}\frac{x^{2n+2}}{2n+3}}{\frac{2^{2n}n!}{(2n)!}\frac{x^{2n}}{2n+1}}\\ &=(2x)^2\frac{2n+1}{2n+3}\frac{(n+1)!}{n!}\frac{(2n)!}{(2n+2)!}\\ &=(2x)^2\frac{2n+1}{2n+3}\frac{n+1}{(2n+2)(2n+1)}\\ &=\frac{n+1}{n+3/2}\frac{x^2}{n+1}, \end{align}$$ so we may write $$t_n=\frac{(1)_n}{(3/2)_n}\frac{(x^2)^n}{n!}t_0,$$ where $$(z)_n=\frac{\Gamma(z+n)}{\Gamma(z)}=z(z+1)(z+2)\cdots(z+n-1).$$ Since $t_0=1$, we have $$\sum_{n\ge0}t_n=\,_1F_1(1;\tfrac32;x^2),$$ and thus $$f(x)=x\,_1F_1(1;\tfrac32;x^2),$$ where $$_pF_q(a_1,...,a_p;b_1,...,b_q;z)=\sum_{n\ge0}\frac{(a_1)_n\cdots(a_p)_n}{(b_1)_n\cdots(b_q)_n}\frac{z^n}{n!}$$ is the generalized hypergeometric function.

Anyway, for $\Re(b)>\Re(a)>0$, we have the integral representation $$_1F_1(a;b;z)=\frac{\Gamma(b)}{\Gamma(a)\Gamma(b-a)}\int_0^1e^{zt}t^{a-1}(1-t)^{b-a-1}dt,$$ from here. Using this, $$f(x)=\frac{x\Gamma(\tfrac32)}{\Gamma(1)\Gamma(\tfrac12)}\int_0^1\frac{e^{x^2 t}}{\sqrt{1-t}}dt=\frac{x}{2}\int_0^1\frac{e^{x^2t}}{\sqrt{1-t}}dt.$$ Then write $z=\sqrt{1-t}$, so that $-2dz=\frac{dt}{\sqrt{1-t}}$, and $$f(x)=x\int_0^1e^{x^2(1-z^2)}dz=xe^{x^2}\int_0^1e^{-(xz)^2}dz.$$ Finally, writing $u=xz$, we have $$f(x)=e^{x^2}\int_0^{x}e^{-u^2}du.$$ Recalling the definition of the Error Function $\mathrm{erf}(z)$, we have $$f(x)=\frac{\sqrt\pi}2e^{x^2}\mathrm{erf}(x).$$ I hope this helps.

Answer 3

Note that $$ \prod _{i=1}^n\left(2i-1\right)=(2n-1)!!\tag1 $$ Therefore, the sum you are looking for is $$ \sum_{n=1}^\infty\frac{2^{n-1}x^{2n-1}}{(2n-1)!!}\tag2 $$ For $u\ge0$, let $$ f(u)=\sum_{n=1}^\infty\frac{(2u)^{n-\frac12}}{(2n-1)!!}\tag3 $$ Then $$ \begin{align} f'(u) &=\frac{\mathrm{d}}{\mathrm{d}u}\sum_{n=1}^\infty\frac{(2u)^{n-\frac12}}{(2n-1)!!}\tag{4a}\\ &=\sum_{n=1}^\infty\frac{(2u)^{n-\frac32}}{(2n-3)!!}\tag{4b}\\ &=\frac1{\sqrt{2u}}+\sum_{n=1}^\infty\frac{(2u)^{n-\frac12}}{(2n-1)!!}\tag{4c}\\[3pt] &=\frac1{\sqrt{2u}}+f(u)\tag{4d} \end{align} $$ Explanation:
$\text{(4a)}$: differentiate $(3)$
$\text{(4b)}$: differentiate term by term
$\text{(4c)}$: pull the $n=1$ term out front and re-index
$\text{(4d)}$: apply $(3)$

Therefore, rearranging $(4)$ and multiplying by the integrating factor of $e^{-u}$, we get $$ \left(e^{-u}f(u)\right)'=e^{-u}f'(u)-e^{-u}f(u)=\frac{e^{-u}}{\sqrt{2u}}\tag5 $$ Integrating, noting that $f(0)=0$, yields $$ \begin{align} e^{-u}f(u) &=\int_0^u\frac{e^{-t}}{\sqrt{2t}}\,\mathrm{d}t\tag{6a}\\ &=\sqrt2\int_0^{\sqrt{u}}e^{-t^2}\,\mathrm{d}t\tag{6b}\\[6pt] &=\sqrt{\pi/2}\operatorname{erf}\left(\sqrt{u}\right)\tag{6c} \end{align} $$ Explanation:
$\text{(6a)}$: integrate $(5)$
$\text{(6b)}$: substitute $t\mapsto t^2$
$\text{(6c)}$: apply the definition of $\operatorname{erf}$

Thus, $(6)$ yields $$ f(u)=e^u\sqrt{\pi/2}\operatorname{erf}\left(\sqrt{u}\right)\tag7 $$ Therefore, we can apply $(3)$ and $(7)$ to get $$ \begin{align} \sum_{k=1}^\infty\frac{2^{n-1}x^{2n-1}}{(2n-1)!!} &=\frac1{\sqrt2}f\!\left(x^2\right)\tag{8a}\\ &=\frac{\sqrt\pi}2e^{x^2}\operatorname{erf}(x)\tag{8b} \end{align} $$ Since the series and $\operatorname{erf}(x)$ are odd functions, we can remove the positivity restriction.