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.
Best Answer
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.