# Is it possibile to obtain the sum of the following series without using hypergeometric functions?

I know that the following series: $$\sum_{n=1}^{+\infty}\frac{ (n!)^2}{(2n)!}$$ converges. If I plug it in Wolphram Alpha, I can see that its sum is $$\frac{1}{27} \left(9 + 2 \sqrt{3} \pi\right).$$ Is it possibile to obtain it without the use of the hypergeometric functions?

Using geometric series (and their derivatives), we get $$\sum_{n=0}^\infty(2n+1)x^n=\frac{1+x}{(1-x)^2}\tag1$$ Let $$\alpha=\frac{1+i\sqrt3}2$$, then \begin{align} \sum_{n=0}^\infty\frac{n!^2}{(2n)!} &=\sum_{n=0}^\infty(2n+1)\int_0^1t^n(1-t)^n\,\mathrm{d}t\tag2\\ &=\int_0^1\frac{1+t(1-t)}{(1-t(1-t))^2}\,\mathrm{d}t\tag3\\[3pt] &=\int_0^1\frac{1+t-t^2}{\left(1-t+t^2\right)^2}\,\mathrm{d}t\tag4\\ &=\int_0^1\left(-\frac23\frac1{(t-\alpha)^2}-\frac23\frac1{(t-\bar\alpha)^2}+i\frac1{3\sqrt3}\left(\frac1{t-\bar\alpha}-\frac1{t-\alpha}\right)\right)\mathrm{d}t\tag5\\[3pt] &=\left[\color{#C00}{\frac23\frac1{t-\alpha}}+\color{#090}{\frac23\frac1{t-\bar\alpha}}+i\frac1{3\sqrt3}\big(\color{#00F}{\log(t-\bar\alpha)}\color{#C90}{-\log(t-\alpha)}\big)\right]_0^1\tag6\\[6pt] &=\color{#C00}{\frac23}+\color{#090}{\frac23}+\color{#00F}{\frac\pi{9\sqrt3}}\color{#C90}{+\frac\pi{9\sqrt3}}\tag7\\[6pt] &=\frac43+\frac{2\pi}{9\sqrt3}\tag8 \end{align} Explanation:
$$(2)$$: Beta Function
$$(3)$$: apply $$(1)$$
$$(4)$$: expand
$$(5)$$: Partial Fractions
$$(6)$$: integrate each term
$$(7)$$: evaluate at limits
$$(8)$$: combine

Subtracting the $$n=0$$ term, we get $$\sum_{n=1}^\infty\frac{n!^2}{(2n)!}=\frac13+\frac{2\pi}{9\sqrt3}\tag9$$

• Nice answer. (+1) – Markus Scheuer Dec 15 '18 at 21:00

I started from the fact that

$$\Gamma(n+\frac{1}{2})=\frac{(2n)!}{4^n n!}\sqrt{\pi}$$ } Divide both side by $$n!$$ and express $$\frac{(n!)^2}{(2n)!}$$

We get

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

Introduce $$\beta$$ function and take the sum of both sides from $$0$$ to $$\infty$$

$$S=\sum\limits_{n=1}^\infty\frac{(n!)^2}{(2n)!}=\sum\limits_{n=0}^\infty \frac{2n+1}{2^{2n+1}}\beta(n+1,\frac{1}{2})-1$$

Based on the definition of $$\beta$$ function can be written

$$S=\sum\limits_{n=0}^\infty \frac{2n+1}{2^{2n+1}}\int\limits_0^1\frac{t^n}{ \sqrt{1-t}} dt-1=\int\limits_0^1\sum\limits_{n=0}^\infty\frac{2n+1}{2^{2n+1}}\frac{t^n}{ \sqrt{1-t}}dt-1$$

The integral can be devided into two parts:

$$\int\limits_0^1\frac{1}{\sqrt{1-t}}\sum\limits_{n=0}^\infty n(\frac{t}{4})^n dt+\frac{1}{2}\int\limits_0^1\frac{1}{\sqrt{1-t}}\sum\limits_{n=0}^\infty (\frac{t}{4})^n dt-1$$

Take $$\frac{t}{4}\sum\limits_{n=0}^\infty n(\frac{t}{4})^{n-1}=t\frac{d}{dt}\sum\limits_{n=0}^\infty (\frac{t}{4})^n=\frac{4t}{(4-t)^2}$$ and $$\sum\limits_{n=0}^\infty (\frac{t}{4})^n=\frac{4}{4-t}$$ into account we get:

$$S=\int\limits_0^1\frac{2(4+t)}{\sqrt{1-t}(4-t)^2}dt-1$$

Let $$x=\sqrt{1-t}$$ then $$S=4\int\limits_0^1 \frac{(5-x^2)}{(3+x^2)}dx-1$$

Forming the integral in the following way:

$$S=\frac{8}{9}\int\limits_0^1\frac{1}{(1+(\frac{x}{\sqrt3})^2)^2}dx+\frac{4}{3}\int\limits_0^1\frac{1-(\frac{x}{\sqrt3})^2}{(1+(\frac{x}{\sqrt3})^2)^2}dx-1$$

Applying the following substitution:$$\frac{x}{\sqrt3}=\tan \theta$$ we receie:

$$S=\frac{8\sqrt3}{9}\int\limits_0^{\frac{\pi}{6}}\cos^2 \theta d\theta+\frac{4\sqrt3}{3}\int\limits_0^{\frac{\pi}{6}}\cos2\theta d\theta-1=\frac{2\sqrt3\pi}{27}+\frac{1}{3}$$

• Thank you very much for all. – JV.Stalker Dec 15 '18 at 21:15

If you also accept non-senseful alternatives:

$$\displaystyle\sum\limits_{n=1}^\infty\frac{n!^2}{(2n)!} = \sum\limits_{n=1}^\infty\frac{1}{(2n)!} \int\limits_0^\infty\frac{t^n}{e^t}dt \int\limits_0^\infty\frac{s^n}{e^s}ds = \int\limits_0^\infty \int\limits_0^\infty \frac{\cosh(\sqrt{ts})-1}{e^{t+s}} dt ds$$

$$\displaystyle \int\limits_0^\infty \frac{\cosh(\sqrt{ts})-1}{e^s} dt = \frac{\sqrt{\pi}}{2}e^{t/4}\sqrt{t}~\text{erf}\left(\frac{\sqrt{t}}{2}\right)$$

$$\displaystyle a>0 :\enspace \int\limits_0^\infty \frac{\sqrt{t}~\text{erf}\left(\frac{\sqrt{t}}{2}\right)}{e^{at}} dt = \frac{\frac{2\sqrt{a}}{4a+1} + \cot^{-1}(2\sqrt{a})}{\sqrt{\pi}a^{3/2}}\enspace$$ ; $$~$$ here: $$\enspace\displaystyle a:=\frac{3}{4}$$

see e.g. erf(x)

• The differential in the second integral should be $ds$, I think. – zar Dec 14 '18 at 12:28
• @zar: Of course, thank you ! Corrected. – user90369 Dec 16 '18 at 12:05