TITLE: Calculate the following limit using the Central Limit Theorem
QUESTION [6 upvotes]: A question about the CLT

Using Central Limit Theorem to show that
$$\lim_{n \to \infty} \frac{8^n}{27^n} \sum_{k=0}^n \binom{3n}{k}\frac{1}{2^k}=0$$

I have tried to define a sequence ${X_n}$ with $X_n$~$Bi(3,p)$,and then put $S_n=X_1+ \ldots + X_n$ then clearly $S_n$~$Bi(3n,p)$ but I have not been able to find the right $p$, and I am not sure how to use CLT, any advice would be much appreciated.

REPLY [6 votes]: Hint:
$$\frac{8^n}{27^n}\cdot\frac{1}{2^k}=\left( \frac{1}{3} \right)^k\cdot\left(\frac{2}{3}  \right)^{3n-k}$$
thus your pmf is
$$\mathbb{P}[X=k]=\binom{3n}{k}\left( \frac{1}{3} \right)^k\cdot\left(\frac{2}{3}  \right)^{3n-k}$$
that is $X\sim \text{Bin}\left(3n;\frac{1}{3} \right)$
with mean $\mathbb{E}[X]=n$ and variance $\mathbb{V}[X]=\frac{2}{3}n$