Sum: $\sum\limits_{n=0}^\infty \frac{n!}{(2n)!}$

Question

I'm struggling with the following sum:

$$\sum_{n=0}^\infty \frac{n!}{(2n)!}$$

I know that the final result will use the error function, but will not use any other non-elementary functions. I'm fairly sure that it doesn't telescope, and I'm not even sure how to get $\operatorname {erf}$ out of that.

Can somebody please give me a hint? No full answers, please.

Answer 1

One might note that:

$$\frac{n!}{(2n)!}=\frac{\sqrt\pi}{4^n\Gamma(n+1/2)}$$

Indeed, this makes your series a special case of the following function:

$$f_\alpha(x)=\sum_{n=0}^\infty\frac{x^n}{\Gamma(n+\alpha)}$$

With $S=\sqrt\pi f_{1/2}(1/4)$.

Wonderfully, this problem serves as a good introduction for fractional derivatives. Let us define here the following:

$$\frac{\mathrm d^\alpha}{\mathrm dx^\alpha}f(x)=\frac1{\Gamma(1-\{\alpha\})}\frac{\mathrm d^{\lceil\alpha\rceil}}{\mathrm dx^{\lceil\alpha\rceil}}\int_0^x(x-t)^{-\{\alpha\}}f(t)~\mathrm dt$$

where $\lceil\alpha\rceil$ is the ceiling function and $\{\alpha\}=\alpha-\lfloor\alpha\rfloor$ is the fractional part of $\alpha$. From the above definition, one can deduce that

$$\frac{\mathrm d^\alpha}{\mathrm dx^\alpha}x^\beta=\frac{x^{\beta-\alpha}\Gamma(\beta+1)}{\Gamma(\beta-\alpha+1)}$$

From this, one may deduce that

$$\frac{x^n}{\Gamma(n+\alpha)}=x^{1-\alpha}\frac{\mathrm d^{1-\alpha}}{\mathrm dx^{1-\alpha}}\frac{x^n}{n!}$$

Thus, we find that

$$f_\alpha(x)=x^{1-\alpha}\frac{\mathrm d^{1-\alpha}}{\mathrm dx^{1-\alpha}}e^x$$

In the particular case of $\alpha=1/2$,

$$\frac{\mathrm d^{1/2}}{\mathrm dx^{1/2}}e^x=\frac1{\Gamma(1/2)}\int_0^x(x-t)^{-1/2}e^t~\mathrm dt$$

Let $x-t=u^2$ to get

$$\frac{\mathrm d^{1/2}}{\mathrm dx^{1/2}}e^x=\frac1{\Gamma(3/2)}\int_0^{\sqrt x}u^2e^{x-u^2}~\mathrm du$$

Whereupon the relationship to the error function is immediate.

The general case for $f_\alpha(x)$ is left as extra credit for the reader with the additional hint that incomplete Gamma functions are recommended.

Answer 2

Here's a hint. Write the fraction as $\int_0^\infty\frac{x^n}{(2n)!}e^{-x}dx$. The sum is then $\int_0^\infty\cosh\sqrt{x}e^{-x} \, dx=\int_0^\infty 2y\cosh y e^{-y^2} \, dy$. Can you take it from there?

Answer 3

The series can be written also in terms of the Incomplete Gamma function. As noted by Simply Beautiful Art we have $$\sum_{n\geq0}\frac{n!}{\left(2n\right)!}=\sum_{n\geq0}\frac{\Gamma\left(1/2\right)}{4^{n}\Gamma\left(n+1/2\right)}$$ $$=1+\frac{1}{4}\sum_{n\geq0}\frac{\Gamma\left(1/2\right)}{4^{n}\Gamma\left(n+1+1/2\right)}=1+\frac{1}{4}\sum_{n\geq0}\frac{1}{4^{n}\left(1/2\right)_{n+1}}$$ where $\left(x\right)_{n}=\Gamma\left(x\right)/\Gamma\left(x+n\right)$ is the Pochhammer symbol and since $$\gamma\left(a,z\right)e^{z}z^{-a}=\sum_{n\geq0}\frac{z^{n}}{\left(a\right)_{n+1}},\,a\neq-k,\,k\in\mathbb{N}$$ we have $$\sum_{n\geq0}\frac{n!}{\left(2n\right)!}=\color{red}{1+\frac{e^{1/4}\gamma\left(1/2,1/4\right)}{2}}\approx \color{blue}{1.5923}$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 0}^{\infty}{n! \over \pars{2n}!} & = 1 + \sum_{n = 1}^{\infty}{\Gamma\pars{n} \over \pars{n - 1}!}\, {\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}} \\[5mm] & = 1 + \sum_{n = 0}^{\infty}{1 \over n!}\,\ \overbrace{% {\Gamma\pars{n + 1}\Gamma\pars{n + 2} \over \Gamma\pars{2n + 3}}} ^{\ds{\substack{\ds{=\ \mrm{B}\pars{n + 1,n + 2}.}\\[1mm] \ds{\mrm{B}:\ Beta\ Function}}}} \\[5mm] & = 1 + \sum_{n = 0}^{\infty}{1 \over n!}\ \overbrace{% \int_{0}^{1}t^{n}\pars{1 - t}^{n + 1}\,\dd t}^{\ds{\mrm{B}\pars{n + 1,n + 2}}} \\[5mm] & = 1 + \int_{0}^{1}\pars{1 - t}\sum_{n = 0}^{\infty} {\bracks{t\pars{1 - t}}^{\,n} \over n!}\,\dd t \\[5mm] & = 1 + \int_{0}^{1}\pars{1 - t}\expo{t\,\pars{1 - t}}\,\dd t \\[5mm] & = 1 + \int_{-1/2}^{1/2}\pars{{1 \over 2} - t}\exp\pars{{1 \over 4} - t^{2}}\,\dd t \\[5mm] & = 1 + \expo{1/4}\ \overbrace{\int_{0}^{1/2}\expo{-t^{2}}\,\dd t} ^{\ds{{1 \over 2}\,\root{\pi}\,\mrm{erf}\pars{1 \over 2}}}\qquad \pars{~\mrm{erf}:\ Error\ Function~} \\[5mm] & = \bbx{1 + {1 \over 2}\,\root{\pi}\expo{1/4}\,\mrm{erf}\pars{1 \over 2}} \approx 1.5923 \end{align}