Ordinary generating function for $\binom{3n}{n}$

Question

The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}} \, , \quad |x| < \frac{1}{4},$$

can be derived by using the duplication formula for the gamma function and the generalized binomial theorem.

But what about the ordinary generating function for $ \displaystyle \binom{3n}{n}$?

According to Wolfram Alpha, $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{2\cos \left(\frac{1}{3} \arcsin \left(\frac{3 \sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \, , \quad |x| < \frac{4}{27}. $$

Any suggestions on how to prove this?

EDIT:

Approaching this problem using the fact that $$ \text{Res} \Big[ \frac{(1+z)^{3n}}{z^{n+1}},0 \Big] = \binom{3n}{n},$$

I get $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = -\frac{1}{2 \pi i x} \int_{C} \frac{dz}{z^{3}+3z^{2}+3z - \frac{z}{x}+1},$$

where $C$ is a circle centered at $z=0$ such that every point on the circle satisfies $ \displaystyle\Big|\frac{x(1+z)^{3}}{z} \Big| < 1$.

Evaluating that contour integral would appear to be quite difficult.

Answer 1

Standard conversion to hypergeometric series and use of the duplication and triplication formulas for the $\Gamma$-function yields \begin{equation} \sum_{n\ge 0}\binom{3n}{n}z^n = \sum_{n\ge 0}\frac{\Gamma(3n+1)}{\Gamma(2n+1)}\frac{z^n}{n!} = \sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(n+1)} {\Gamma(n+1/2)\Gamma(n+1)}\frac{z^n}{n!} \frac{(2\pi)^{1/2} 3^{3n+1/2}}{2\pi2^{2n+1/2}} \end{equation}

\begin{equation} = \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)} \sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(1/2)} {\Gamma(n+1/2)\Gamma(1/3)\Gamma(2/3)}\frac{z^n}{n!} \frac{(2\pi)^{1/2} 3^{3n+1/2}}{2\pi2^{2n+1/2}} \end{equation}

\begin{equation} = \sqrt{3/(4\pi)} \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)} \sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(1/2)} {\Gamma(n+1/2)\Gamma(1/3)\Gamma(2/3)}\frac{(2^{-2} 3^3z)^n}{n!} \end{equation}

\begin{equation} = \sqrt{3/(4\pi)} \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)} \sum_{n\ge 0}\frac{(1/3)_n (2/3)_n} {(1/2)_n}\frac{(2^{-2} 3^3z)^n}{n!} \end{equation}

\begin{equation} = \sqrt{3/(4\pi)} \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)}{} _2F_1(1/3, 2/3; 1/2; 2^{-2}3^3z) \end{equation} \begin{equation} = \sqrt{3} \frac{\Gamma(1/3)\Gamma(2/3)}{2\pi}{} _2F_1(1/3, 2/3; 1/2; 2^{-2}3^3z) \end{equation} Furthermore by equation 15.1.18 of Abramowitz/Stegun this Gaussian Hypergeometric Function can be reduced by \begin{equation} _2F_1(a,1-a;1/2;\sin^2z)=\frac{\cos[(2a-1)z]}{\cos z} \end{equation} with parameter $a=1/3$. Furthermore $\Gamma(1/3)\Gamma(2/3) = 2\pi/\sqrt{ 3}$ according to OEIS sequence A073006.

Answer 2

Proposition : $$ f(z,a) = \int_{0}^{\infty} \dfrac{x^z}{x^2 +2ax +1} \mathrm{d}x = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))}$$

Proof : Note that,

$$ \sum_{n=0}^{\infty}{{U}_{n}(a) {(-x)}^{n}} = \dfrac{1}{x^2 +2ax+1} $$

where $ U_{n} (x) $ is the Chebyshev Polynomial of the second kind.

$$ \implies \sum_{n=0}^{\infty}{{U}_{n}(a) \Gamma(n+1) \frac{{(-x)}^{n}}{n!}} = \frac{1}{x^2 +2ax +1} $$

Using Ramanujan Master Theorem, we have,

$$ f(z,a) = \dfrac{\pi}{\sin \pi z} U_{-s} (a) $$

$$ = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))} \quad \square $$

Now, using Gamma Triplication Formula,

$$ (3n)! \; = \; \Gamma(3n+1) \; = \; \dfrac{1}{2\pi} 3^{2n+\frac12} \Gamma \left(n+\dfrac13\right)\Gamma \left(n+\dfrac23\right)\Gamma(n+1) $$

and hence

$$ {3n \choose n} \; = \; \frac{3^{3n+\frac12}}{2\pi} \times \frac{\Gamma \left(n+\dfrac13\right)\Gamma \left(n+\dfrac23 \right)}{\Gamma(2n+1)} \; = \;\frac{3^{2n+\frac12}}{2\pi} \operatorname{B} \left(n+\dfrac23,n+\dfrac13\right) $$

Thus,

$$ \begin{array}{rcl}\displaystyle \text{S} \; =\; \sum_{n=0}^\infty {3n \choose n}x^n & = & \displaystyle\frac{\sqrt{3}}{2\pi}\sum_{n=0}^\infty 3^{2n} x^n \int_0^1 u^{n-\frac13}(1-u)^{n-\frac23}\,du \\ & = & \displaystyle \frac{\sqrt{3}}{2\pi} \int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 27xu(1-u)}\,du \; =\; \frac{\sqrt{3}}{2\pi}\int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 4a^2\, u(1-u)}\,du \end{array} $$

where $a = \dfrac{3}{2} \sqrt{3x} $. Substituting $u = \sin^2\theta$ and then $\tan\theta = t$, we have,

$$\text{S} = \dfrac{\sqrt{3}}{2 \pi} \int_0^\infty \frac{t^{\frac{1}{3}}(1 + t^2)}{(t^2 + 2at + 1)(t^2 - 2at + 1)} \mathrm{d}t $$

Using Partial Fraction and the Proposition, we have,

$$\text{S} = \dfrac{\sqrt{3}}{2\pi} \left[ \dfrac{1}{2} f \left( \dfrac{1}{3} , -a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{4}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{10}{3} , -a \right) + \dfrac{1}{2} f \left( \dfrac{1}{3} , a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{4}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{10}{3} , a \right)\right] $$

After simplification, we have,

$$ \text{S} = \dfrac{2\cos \left(\frac{1}{3} \sin^{-1} \left(\dfrac{3\sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \quad \square $$

Answer 3

The ordinary generating function of ${kn \choose n}$ is (the derivative of) a power series expansion for the real root near $1$ of a degree $n$ polynomial that looks very much like $$ x^k - x - t=0.$$ There is a combinatorial interpretation using $k$-generalizations of Catalan numbers.

The fraction in the question therefore comes from solving a cubic equation, and the trigonometric solution is for the case with three real roots. This is consistent with $x^3 - x = t$ for small $t$.

Unfortunately I don't remember the exact polynomial. "Hypergeometric quintic" at Wikipedia finds

http://en.wikipedia.org/wiki/Bring_radical#Series_representation

Answer 4

The following is a proof of the hypergeometric identity $$_2F_1\left(a,1-a;\frac{1}{2};\sin^2(x)\right)=\frac{\cos[(2a-1)x]}{\cos x} \, , \quad - \frac{\pi}{2} < x < \frac{\pi}{2} \tag{1}.$$

This identity is used in R. J. Mathar's answer.

Similar to my answer here, I will first use the generalized binomial theorem to show that $$_2F_1\left(a,1-a;\frac{1}{2};-z^{2}\right) = \frac{1}{2 \sqrt{1+z^{2}}} \left[\left(\sqrt{1+z^{2}}+z \right)^{2a-1} + \left(\sqrt{1+z^{2}}-z \right)^{2a-1} \right] \, , \quad |z| <1. $$

Identity $(1)$ then follows if $z$ is replaced with $i \sin (x)$.

---

\begin{align} &\frac{1}{2\sqrt{1+z^{2}}} \left[\left(\sqrt{1+z^{2}}+z \right)^{2a-1} + \left(\sqrt{1+z^{2}}-z \right)^{2a-1}\right] \\ &= \frac{1}{2\sqrt{1+z^{2}}} \left[\sum_{k=0}^{\infty} \binom{2a-1}{k} z^{k}\left(\sqrt{1+z^{2}} \right)^{2a-1-k} + \sum_{j=0}^{\infty} \binom{2a-1}{j} (-z)^{j} \left(\sqrt{1+z^{2}} \right)^{2a-1-j}\right ] \\ &= \sum_{k=0}^{\infty} \binom{2a-1}{2k} z^{2k} (1+z^{2})^{a-1-k}\\ &= \sum_{k=0}^{\infty} \binom{2a-1}{2k} z^{2k} \sum_{l=0}^{\infty} \binom{a-1-k}{l}z^{2l}\\ & \stackrel{(2)}= \sum_{n=0}^{\infty} \sum_{m=0}^{n}\binom{2a-1}{2m} \binom{a-1-m}{n-m} z^{2n} \\ &=\sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)}\sum_{m=0}^{n} \frac{\Gamma(a-m)}{\Gamma(2m+1)\Gamma(2a-2m) \Gamma(n-m+1)} \, z^{2n} \\ & \stackrel{(3)}= \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \sum_{m=0}^{n} \frac{\pi}{2^{2a-1}} \frac{1}{\Gamma \left(m+\frac{1}{2}\right) \Gamma(m+1) \Gamma(a-m+\frac{1}{2}) \Gamma(n-m+1)} \, z^{2n} \\ &= \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \frac{\pi}{2^{2a-1}} \frac{1}{\Gamma(n+ \frac{1}{2}) \Gamma(a+ \frac{1}{2})} \sum_{m=0}^{n} \binom{n-\frac{1}{2}}{n-m} \binom{a-\frac{1}{2}}{m} \, z^{2n} \\ & \stackrel{(4)}= \sum_{n=0}^{\infty} \frac{\Gamma(2a)}{\Gamma(a-n)} \frac{\pi}{2^{2a-1}}\frac{1}{\Gamma(n+ \frac{1}{2}) \Gamma(a+ \frac{1}{2})} \binom{a+n-1}{n} \, z^{2n} \\ & \stackrel{(5)} = \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(a-n)} \frac{\sqrt{\pi}}{\Gamma (n+\frac{1}{2})} \, \frac{z^{2n}}{n!} \frac{\Gamma(a)}{\Gamma(a)} \\ & \stackrel{(6)} = \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(a)} \frac{\Gamma(1-a+n)}{\Gamma(1-a)} \frac{\Gamma(\frac{1}{2})}{\Gamma(n+ \frac{1}{2})} \frac{(-z^{2})^{n}}{n!} \\ &= \, _2F_{1} \left(a, 1-a; \frac{1}{2}; -z^{2} \right) \end{align}

---

$(2)$: Cauchy product

$(3)$: Duplication formula for the gamma function

$(4)$: Chu-Vandermonde identity

$(5)$: Duplication formula again

$(6)$: $\frac{\Gamma(a)}{\Gamma(a-n)}= (-1)^{n} \frac{\Gamma(1-a+n)}{\Gamma(1-a)}$

---