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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.

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  • $\begingroup$ I think the limit must be $\tfrac 12$ and not zero $\endgroup$
    – Evangelopoulos Foivos
    Commented Dec 17, 2021 at 12:18
  • $\begingroup$ Oh really? Please you can explain me? $\endgroup$
    – Nick Weber
    Commented Dec 17, 2021 at 20:29

1 Answer 1

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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 $\mathbb{V}[X]=\frac{2}{3}n$

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  • $\begingroup$ Thank you very much for your help, I am right now trying to apply CLT, the next step would be to $$\mathbb{P}[X_1+...+X_n \leq n]$$ and aplly CLT? $\endgroup$
    – Nick Weber
    Commented Dec 17, 2021 at 10:47

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