Please help me find the sum of this series:
$$1 + \frac{2}{3}\cdot\frac{1}{2} + \frac{2}{3}\cdot\frac{5}{6}\cdot\frac{1}{2^2} + \frac{2}{3}\cdot\frac{5}{6}\cdot\frac{8}{9}\cdot\frac{1}{2^3} + \cdots$$
All I could figure out was to find the $n^{\text{th}}$ term as:
$$a_n = \frac{2 \cdot (2+3) \cdots(2+3(n-1))}{3 \cdot 6 \cdot 9 \cdots 3(n-1)} \cdot\frac{1}{2^{n-1}}$$
What To do ahead of it. I don't know. Please help.
Let $S$ denote the sum. We write each term (with indices starting at $0$) as
$$ \left( \prod_{k=1}^{n} \frac{3k-1}{3k} \right) \frac{1}{2^n} = \frac{\prod_{k=0}^{n-1} (-\frac{2}{3}-k)}{n!} \left(-\frac{1}{2}\right)^n = \binom{-2/3}{n} \left(-\frac{1}{2}\right)^n. $$
Then we easily recognize $S$ as a bionmial series and hence
$$ S = \sum_{n=0}^{\infty}\binom{-2/3}{n} \left(-\frac{1}{2}\right)^n = \left(1 - \frac{1}{2}\right)^{-2/3} = 2^{2/3}.$$