Limit of a sum (no probabilities)

Question

Show that $$\lim_{n\to+\infty}\left(\frac{2}{3}\right)^n\sum_{k=0}^{[n/3]}\binom{n}{k}2^{-k}=\frac{1}{2}$$ without using probabilities.

$[\;\cdot\;]$ denotes the integer part.