Find the sum: $$\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$$

My try:

I played a bit with the coefficient to make it look easier/familiar:

First attempt: $$\begin{align} \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n &= \sum_{n=0}^\infty \frac{n!}{2^n(2n-1)!!}x^n \\ &= \sum_{n=0}^\infty \frac{n!}{(2n-1)!!}\left(\frac x2\right)^n \end{align}$$ Second attempt: $$\begin{align} \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n &= \sum_{n=0}^\infty \frac{n!\cdot n!}{(n+n)!}x^n \\ &= \sum_{n=0}^\infty \frac{1}{{2n \choose n}}x^n \end{align}$$ However, I could not proceed with any of them. Also, I have figured out that the convergence radius is $4$.

My research:

I have also found the same sum has been discussed at AoPS, which unfortunately uses Beta function that my course has not covered yet.
Entering the sum to Wolphram Alpha, I got the following output for the partial sum:


My background:

As I have already mentioned, I cannot use Gamma, Beta or similar functions. I only know about the convergence theorems on functional series and operations on them. So, I'm looking for some method that uses quite elementary tricks.

Thanks in advance.

  • 3
    $\begingroup$ Hope this helps $$\sum _{n=0}^{\infty } \frac{(n!)^2 x^n}{(2 n)!}=\frac{4 \left(\sqrt{4-x}+\sqrt{x} \arcsin\left(\frac{\sqrt{x}}{2}\right)\right)}{\sqrt{4-x} (4-x)}$$ radius of convergence $r=4$ $\endgroup$
    – Raffaele
    Feb 10, 2021 at 10:07
  • 1
    $\begingroup$ @A-LevelStudent With Mathematica. Maybe it can give some idea to solve it. It's not fair publishing a result as an answer without the proof. $\endgroup$
    – Raffaele
    Feb 10, 2021 at 10:12
  • 6
    $\begingroup$ In equation $(2)$ of this answer, it is shown that $$ \sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}} =\frac4{4-x}\left[1+\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)\right] $$ $\endgroup$
    – robjohn
    Feb 10, 2021 at 10:18
  • 1
    $\begingroup$ Check this: math.stackexchange.com/a/548570. $\endgroup$
    – Martin R
    Feb 10, 2021 at 10:31
  • 1
    $\begingroup$ Many posts with close relatives (here's my own one), but no dedicated question I could see until now ;) $\endgroup$
    – metamorphy
    Feb 10, 2021 at 21:48

2 Answers 2


Suppose that $\displaystyle f(x) = \sum_{n\geqslant 0} a_n x^n$ with $a_0=1$. We shall assume that the series has non-zero radius of convergence. Then, \begin{align*} \frac{1}{2}\frac{d}{dx} \Big( x^2 f(x^2)\Big) = \sum_{n \geqslant 0} (n+1)a_n x^{2n+1} \quad \tag{1}\label{A} \end{align*} and \begin{align*} 2x \frac{d}{dx} \Bigg( \frac{f(x^2)-1}{x} \Bigg) &= \sum_{n \geqslant 1} 2(2n-1)a_n x^{2n-1} \\ &=\sum_{n \geqslant 0}2(2n+1) a_{n+1} x^{2n+1} \tag{2}\label{B} \end{align*} If $f$ now satisfies the differential equation, \begin{align*} 2x \frac{d}{dx} \Bigg( \frac{f(x^2)-1}{x} \Bigg) = \frac{1}{2} \frac{d}{dx} \Big(x^2 f(x^2)\Big) \tag{3}\label{C} \end{align*} we can equate coefficients in the power series \eqref{A} and \eqref{B} to derive, \begin{align*} a_{n+1} = \frac{n+1}{2(2n+1)} a_n, \quad a_0 = 1 \end{align*} which means \begin{align*} a_n = \frac{n!}{2^n(2n-1)(2n-3) \cdots 1} = \frac{(n!)^2}{(2n)!}. \end{align*} and \begin{align*} f(x) = \sum_{n \geqslant 0} \frac{(n!)^2}{(2n)!} x^n. \end{align*} Introduce $\phi(x) = (f(x^2)-1)/x$ so that the differential equation \eqref{C} becomes, \begin{align*} (4-x^2)\phi' -3x \phi = 2 \end{align*} Multiply this by the integrating factor $(4-x^2)^{1/2}$ to obtain, \begin{align*} (4-x^2)^{3/2} \phi' - 3x (4-x^2)^{1/2}\phi=2(4-x^2)^{1/2} \end{align*} which is the same as \begin{align*} \frac{d}{dx} \Big( (4-x^2)^{3/2} \phi \Big) = 2(4-x^2)^{1/2} \tag{4}\label{D}. \end{align*} We note that by construction $\phi(0)=0$.

The right hand side may be integrated using the substitution $x=2\sin\theta$ to give, \begin{align*} \int_0^\xi 2(4-x^2)^{1/2} dx &= 8\int_0^{\sin^{-1}(\xi/2)} \cos^2\theta d\theta \\ &=8 \Big[ \sin\theta\cos\theta \Big]_0^{\sin^{-1}(\xi/2)} + 8\int_0^{\sin^{-1}(\xi/2)} \sin^2\theta d\theta \\ &=8 (\xi/2) (1-\xi^2/4)^{1/2}+8\int_0^{\sin^{-1}(\xi/2)} 1-\cos^2\theta d\theta \\ &= 2 \xi (4-\xi^2)^{1/2} + 8\sin^{-1}(\xi/2) - 8\int_0^{\sin^{-1}(\xi/2)} \cos^2\theta d\theta \end{align*} from which we see, \begin{align*} \int_0^\xi 2(4-x^2)^{1/2} dx = \xi(4-\xi^2)^{1/2} + 4\sin^{-1}(\xi/2). \end{align*} Substituting this in \eqref{D}, and recalling $\phi(0)=0$, we get \begin{align*} \phi(x)(4-x^2)^{3/2} = x(4-x^2)^{1/2} + 4\sin^{-1}(x/2) \end{align*} yielding \begin{align*} f(x) &= \sqrt x \phi(\sqrt x) + 1 \\ &= 1 + \frac{x}{4-x}+\frac{4\sqrt x \sin^{-1}(\frac{\sqrt x }{2})}{(4-x)^{3/2}} \\ &=\frac{4}{4-x}\Bigg(1 + \frac{\sqrt{x} \sin^{-1}(\frac{\sqrt{x}}{2})}{\sqrt{4-x}} \Bigg) \end{align*}


More than tricky ! (and done using a CAS). $$\sum_{n=0}^p \frac{(n!)^2}{(2n)!}x^n=\frac d {dx}\Bigg[\sum_{n=0}^p \frac{(n!)^2}{(n+1)\,(2n)!}x^{n+1}\Bigg]$$ $$\sum_{n=0}^p \frac{(n!)^2}{(n+1)\,(2n)!}x^{n+1}=x \, _3F_2\left(1,1,1;\frac{1}{2},2;\frac{x}{4}\right)+\color{red}{\text{monster}}$$ $$\color{red}{\text{monster}}=\frac{((p+1)!)^2 x^{p+2} \, _3F_2\left(1,p+2,p+2;p+\frac{3}{2},p+3;\frac{x}{4}\right)}{(p+2) (2 (p+1))!}$$ $$\frac d {dx}\Big[x \, _3F_2\left(1,1,1;\frac{1}{2},2;\frac{x}{4}\right)\Big]=\frac{4}{4-x}+\frac{4 \sqrt{x} \sin ^{-1}\left(\frac{\sqrt{x}}{2}\right)}{(4-x)^{\frac 32}}$$ and $$\frac d {dx}\big[\color{red}{\text{monster}}\big]=\frac{x^{p+1} \Gamma (p+2)^2 \, _2F_1\left(1,p+2;p+\frac{3}{2};\frac{x}{4}\right)}{\Gamma (2 p+3)}$$ which, in the radius of convergence, tends very quickly to $0$.

  • 2
    $\begingroup$ Thank you for your attention! But I was looking for a quite elementary approach without $\beta, \Gamma, F$ functions that I cannot use. $\endgroup$
    – NodeJS
    Feb 10, 2021 at 11:54

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