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

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 2

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 $i \sin (x)$.

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\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}

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$(2)$: Cauchy product

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

$(4)$: Chu-Vandermonde identity

$(5)$: Duplication formula again

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

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