A question about Central Limit Theorem and the calculation of a limit.
Solution 1:
Note that
\begin{align*} \left(\frac{8}{27}\right)^n \sum_{k=0}^{n} \binom{3n}{k} \frac{1}{2^k} = \sum_{k=0}^{n} \binom{3n}{k} \left( \frac{1}{3} \right)^k \left( \frac{2}{3} \right)^{3n-k} = \mathbf{P}(S_{3n} \leq n), \end{align*}
where $S_{3n} \sim \operatorname{Bin}(3n, \frac{1}{3})$. So by the CLT,
$$ \mathbf{P}(S_{3n} \leq n) = \mathbf{P}\left(\frac{S_{3n} - \mathbf{E}[S_{3n}]}{\sqrt{\mathbf{Var}(S_{3n})}} \leq 0 \right) \xrightarrow{n\to\infty} \mathbf{P}(Z \leq 0) = \frac{1}{2}, $$
where $Z \sim \mathcal{N}(0, 1)$ is a standard normal variable. I also included the graph of this quantity as a function of $n$:
