7
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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 $$ \sum_{n=1}^\infty\frac{n!^2}{(2n)!}=\frac13+\frac{2\pi}{9\sqrt3}\tag9 $$

$\endgroup$
  • $\begingroup$ Nice answer. (+1) $\endgroup$ – Markus Scheuer Dec 15 '18 at 21:00
4
$\begingroup$

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:

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

$\endgroup$
  • 1
    $\begingroup$ Thank you very much for all. $\endgroup$ – JV.Stalker Dec 15 '18 at 21:15
1
$\begingroup$

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)

$\endgroup$
  • 1
    $\begingroup$ The differential in the second integral should be $ds$, I think. $\endgroup$ – zar Dec 14 '18 at 12:28
  • 1
    $\begingroup$ @zar: Of course, thank you ! Corrected. $\endgroup$ – user90369 Dec 16 '18 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.