Challenge: Prove $\displaystyle \sum_{n \in \mathbb{N}} \frac{n!}{(2n+1)!}=e^{1/4}\sqrt{\pi}\ \ \mathrm{erf}(\frac{1}{2})$

Question

I stumbled upon this cute sum while messing about, and I want to see what other solutions people propose before I put forward my own (which may be unnecessarily complicated).

You can use any maths you like.

NB: $\displaystyle \mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$

Answer 1

Recall for all $x > 0$, $$\sinh x = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!} \quad\implies\quad \frac{\sinh \sqrt{x}}{\sqrt{x}} = \sum_{n=0}^\infty \frac{x^n}{(2n+1)!} $$ and the integral representation of $n!$ $$n! = \int_0^\infty x^n e^{-x}dx$$

The sum at hand can be rewritten as $$ \sum_{n=0}^\infty \int_0^\infty\frac{x^{n}}{(2n+1)!} e^{-x}dx = \int_0^\infty e^{-x}\left(\sum_{n=0}^\infty \frac{x^n}{(2n+1)!}\right)dx = \int_0^\infty e^{-x}\frac{\sinh\sqrt{x}}{\sqrt{x}}dx $$ Change variable to $t = \sqrt{x}$, this reduces to $$\int_0^\infty e^{-t^2}(e^t - e^{-t}) dt = e^{\frac14}\int_0^\infty \left(e^{-(t-\frac12)^2} - e^{-(t+\frac12)^2}\right)dt = e^{\frac14}\left(\int_{-\frac12}^\infty - \int_{\frac12}^\infty\right) e^{-t^2}dt\\ = e^{\frac14}\int_{-\frac12}^{\frac12} e^{-t^2}dt = 2e^{\frac14}\int_0^{\frac12} e^{-t^2}dt = e^{\frac14}\sqrt{\pi}\,\mathrm{erf}\left(\frac12\right) $$

Answer 2

Outline:

• We know that $$e^{1/4} = \sum_{n=0}^\infty \frac{1}{2^{2n} n!}$$

• and that $$\begin{align} \sqrt{\pi}\operatorname{erf}\left(\frac{1}{2}\right) &= \int_0^{1/2} e^{-t^2}dt = \int_0^{1/2} \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!} dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\int_0^{1/2} t^{2n} dt \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n}{n! 2^{2n}} \frac{1}{2n+1} \end{align}$$

so that, writing it as a product of series and using the Cauchy product of these, $$\begin{align} \sqrt{\pi}\operatorname{erf}\left(\frac{1}{2}\right) e^{1/4} &= \sum_{n=0}^{\infty} \frac{(-1)^n}{n! 2^{2n}} \frac{1}{2n+1}\sum_{n=0}^\infty \frac{1}{2^{2n} n!} = \sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^k}{k! 2^{2k} (2k+1)}\frac{1}{(n-k)!2^{2n-2k}} \\ &= \sum_{n=0}^\infty \frac{1}{2^{2n}n!}\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{k! (2k+1)} = \sum_{n=0}^\infty \frac{1}{2^{2n}n!}\frac{(2n)!!}{(2n+1)!!} \end{align}$$ and from there it is not hard to conclude from the properties of the double factorial.

(If needed, I can fill in the remaining steps later.)

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Finishing it: $$\begin{align} \frac{(2n)!!}{(2n+1)!!} = \frac{2^n n!}{(2n+1)!!} = \frac{2^n n!\sqrt{\pi}}{2^{n+1}\Gamma(\frac{1}{2}+n+1)} \end{align}$$ from which, using the properties of the Gamma function, $$\begin{align} \frac{(2n)!!}{2^{2n}n!(2n+1)!!} &= \frac{\sqrt{\pi}}{2^{2n+1}\Gamma(\frac{1}{2}+n+1)} = \frac{\sqrt{\pi}2^{2n+2}(n+1)!}{2^{2n+1}\sqrt{\pi}(2n+2)!}\\ &= \frac{2 (n+1)n!}{2(n+1)(2n+1)!} = \frac{n!}{(2n+1)!} \end{align}$$ and we are done.

Answer 3

Let $\displaystyle I_n= \int_0^{\infty} x^ne^{-x^2}dx$.

Let $S=\displaystyle \sum_{n \geq 0} \frac{I_n}{n!}=\int_0^{\infty}e^xe^{-x^2}dx=e^{1/4}\int_0^{\infty} e^{-(x-1/2)^2}dx=\frac{1}{2}e^{1/4}\sqrt{\pi}(\mathrm{erf}(\frac{1}{2})+1)$.

Splitting $S$ into odd and even parts, have

$\displaystyle S_{\mathrm{even}}=\sum_{n \geq 0} \int_0^{\infty} \frac{x^{2n}}{2n!}e^{-x^2}dx=\frac{1}{2}\int_0^{\infty} (e^{x}+e^{-x})e^{-x^2}dx=\frac{1}{2}e^{1/4}\sqrt{\pi}$

So $\displaystyle S_{\mathrm{odd}}=\frac{1}{2} e^{1/4}\sqrt{\pi}\ \ \mathrm{erf}(\frac{1}{2})$.

Conclude using $\displaystyle I_{2n+1}=\frac{n!}{2}$ (from integration by parts).

We get $\displaystyle \sum_{n \geq 0} \frac{n!}{(2n+1)!}=e^{1/4}\sqrt{\pi}\ \ \mathrm{erf}(\frac{1}{2})$.

Answer 4

Simply, by using Euler's beta function, $$S=\sum_{n\geq 0}\frac{n!}{(2n+1)!}=\sum_{n\geq 0}\frac{B(n+1,n+1)}{n!}=\int_{0}^{1}\sum_{n\geq 0}\frac{x^n(1-x)^n}{n!}\,dx = \int_{0}^{1}e^{x(1-x)}\,dx$$ and by setting $x=t+\frac{1}{2}$ $$ S = \int_{0}^{1}e^{x(1-x)}\,dx = 2e^{1/4} \int_{0}^{1/2}e^{-t^2}\,dt.$$