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Use the central limit theorem to show that for $x>0$, $$\lim_{n \rightarrow \infty} \frac{1}{3^n} \sum_{k:|3k-2n| \leq \sqrt{2n}x} \binom{n}{k} 2^k = \int^{x}_{-x} \frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}du.$$

Ok so I know $\int^{x}_{-x} \frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}du = \Phi(x)-\Phi(-x)$. I also know the CLT is, if we let $X_1, X_2,...,X_n$ denote the observations of a random sample from a distribution that has mean $\mu$ and variance $\sigma^2$. Then the random variable $Y_n = (\sum^{n}_{i=1}X_i-n\mu)/\sqrt{n}\sigma$ converges in distribution to a random variable which has a normal distribution with mean zero and variance 1. I expanded the RHS for a small $n$, $n=4$ for example. The number of $k$ terms that we can accept depends on how big we make $x$. I am confused on how I am supposed to manipulating the RHS to put in a form for which we could use the CLT. Can any one help me?

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    $\begingroup$ $\phi$ is more commonly used for the standard normal density; $\Phi$ is more often used for the cdf. $\endgroup$ Commented Mar 13, 2016 at 9:11

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The binomial distribution of size $n$ and probability $p$ has probability mass function $P(Y_n=k)=\binom{n}{k}p^k(1-p)^{n-k}$. Setting $p=2/3$ gives:

$$P(Y_n=k)=\binom{n}{k}\left(\frac{2}{3}\right)^k\left(\frac{1}{3}\right)^{n-k}=\frac{1}{3^n}\binom{n}{k}2^k.$$

It follows that the left-hand-side of your equation sneakily represents the probability:

$$\lim_{n\rightarrow\infty}P(|3Y_n-2n|\leq\sqrt{2n}x).$$

$Y_n$ has mean $\frac{2}{3}n$. Do you now see how to apply the CLT (after a few manipulations)?

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