Show $\frac 1 3 C_0+\left(\frac 2 3 \right)\left(\frac 1 3 \right)^2 C_1+\left(\frac 2 3\right)^2 \left(\frac 1 3 \right)^3 C_2 +\dots =\frac 1 2$

Question

From a puzzle involving Markov chains I have the following expression, and calculating $50$ terms of the sum I strongly believe $$\frac 1 3 C_0 + \left(\frac 2 3 \right) \left(\frac 1 3 \right)^2 C_1 + \left(\frac 2 3 \right)^2 \left(\frac 1 3 \right)^3 C_2 + \dots = \frac 1 2$$

where $C_n$ is the nth Catalan number. I know each term of the sum $$x_{n+1} = \left(\frac 2 9 \right) \frac{2(2n+1)}{n+2} x_n$$ so the sum converges, but I don't know how to prove it converges to $1/2$.

Answer 1

Use the generating function for the Catalan numbers: $$c(x) = \sum_{n=0}^\infty C_n x^n = \frac{2}{1+\sqrt{1-4x}}.$$ So, $$\begin{align}\frac 1 3 C_0 + \left(\frac 2 3 \right) \left(\frac 1 3 \right)^2 C_1 + \left(\frac 2 3 \right)^2 \left(\frac 1 3 \right)^3 C_2 + \dots &= \frac13 \sum_{n=0}^\infty C_n \left(\frac29\right)^n \\ &= \frac{1}{3}\cdot \frac{2}{1+\sqrt{1-\frac89}} \\ &= \frac{1}{3} \frac{2}{\frac43} \\&= \frac12.\end{align}$$