$\sum_{n=0}^{\infty}\left ( \frac{2}{27} \right )^n \binom{3n}{n}$

Evaluate the series:
$$\sum_{n=0}^{\infty}\left ( \frac{2}{27} \right )^n \binom{3n}{n}$$

Answer: $\displaystyle \frac{\sqrt{3}+1}{2}$.

Proof: 



We begin by the integral representation of the binomial coefficient which is:
$$2\pi i \binom{3n}{n}=\oint_{\gamma}\frac{\left ( 1+z \right )^{3n}}{z^{n+1}}\,dz$$

whereas $\gamma$ is any circle contour about the origin and radius $\epsilon$.

Our sum is transformed to
$$\sum_{n=0}^{\infty}\left ( \frac{2}{27} \right )^n \frac{\left ( 1+z \right )^{3n}}{z^{n+1}}=\sum_{n=0}^{\infty}\frac{1}{z}\left ( \frac{2\left ( 1+z \right )^3}{27z} \right )^n =\cdots=-\frac{27}{\left ( z-2 \right )\left ( 2z^2+10z-1\right )}$$

because it is a geometric series and converges if-f $\displaystyle \left | \frac{2\left ( 1+z \right )^3}{27z} \right |<1$.

Hence by integrating around the unit circle (it does its work) we have:


$$2\pi i \sum_{n=0}^{\infty}\left ( \frac{2}{27} \right )^n \binom{3n}{n} =\oint_{\left | z \right |=1}-\frac{27}{\left ( z-2 \right )\left ( 2z^2+10z-1 \right )}\,{\rm d}z \Leftrightarrow$$
$$ \Leftrightarrow 2\pi i \sum_{n=0}^{\infty}\left ( \frac{2}{27} \right )^n \binom{3n}{n}=2\pi i \frac{\sqrt{3}+1}{2}\Leftrightarrow \sum_{n=0}^{\infty}\left ( \frac{2}{27} \right )^n \binom{3n}{n}=\frac{\sqrt{3}+1}{2}$$

since there is only one pole within the  circle and the exercise comes to an end.