Proving that the de Moivre-Laplace Theorem is a special case of the Central Limit Theorem. I have a problem that I need help on.
I want to prove that the de Moivre-Laplace Theorem is a special case of the Central Limit Theorem (note that a binomial random variable is a sum of independent Bernoulli trials). This is what I have tried:
Let $ X_{1},X_{2},\ldots,X_{n} $ be Bernoulli trials. If $ S_{n} \stackrel{\text{df}}{=} X_{1} + X_{2} + \cdots + X_{n} $, then
$$
\mathbf{E}[S_{n}] = n \mu \qquad \text{and} \qquad
\mathbf{Var}[S_{n}] = n \sigma^{2},
$$
where $ \mu $ and $ \sigma $ are, respectively, the mean and standard deviation of the $ X_{i} $’s.
Standardizing $ S_{n} $, we get $ Z_{n} = \dfrac{S_{n} - n \mu}{\sqrt{n} \sigma} $. However,
$$
\mathbf{E}[S_{n}] = n p \qquad \text{and} \qquad \mathbf{Var}[S_{n}] = n p q,
$$
so $ Z_{n} = \dfrac{S_{n} - n p}{\sqrt{n p q}} $. Applying the Central Limit Theorem, I get the de Moivre-Laplace Theorem.

Thanks for your help, and have a nice day.
 A: Here is a more rigorous answer than the one offered by nbubis.
The Central Limit Theorem states that for a probability space $ (\Omega,\Sigma,\mathsf{P}) $ and a sequence $ (X_{k})_{k \in \mathbb{N}} $ of i.i.d. random variables on $ (\Omega,\Sigma,\mathsf{P}) $ with mean $ \mu $ and finite standard deviation $ \sigma $, we have
$$
\forall z \in \mathbb{R}: \qquad
  \lim_{n \to \infty}
  \mathsf{P} \!
  \left(
  \frac{\displaystyle \left[ \frac{1}{n} \sum_{k = 1}^{n} X_{k} \right] - \mu}
       {\dfrac{\sigma}{\sqrt{n}}}
  \leq z
  \right)
= \Phi(z),
$$
where $ \Phi $ denotes the standard normal c.d.f.
Now, suppose that $ (X_{k})_{k \in \mathbb{N}} $ is a Bernoulli process on $ (\Omega,\Sigma,\mathsf{P}) $ with probability of success $ p $. Then
$$
\mu = p \qquad \text{and} \qquad \sigma = \sqrt{p (1 - p)}.
$$
Next, define a sequence $ (S_{n})_{n \in \mathbb{N}} $ of random variables on $ (\Omega,\Sigma,\mathsf{P}) $ by
$$
\forall n \in \mathbb{N}: \qquad
S_{n} \stackrel{\text{df}}{=} \sum_{k = 1}^{n} X_{k}.
$$
Then $ S_{n} \sim \operatorname{Binom}(n,p) $ for each $ n \in \mathbb{N} $, and the Central Limit Theorem yields
$$
\forall z \in \mathbb{R}: \qquad
  \Phi(z)
= \lim_{n \to \infty}
  \mathsf{P} \!
  \left( \frac{\dfrac{S_{n}}{n} - p}{\sqrt{\dfrac{p (1 - p)}{n}}} \leq z \right)
= \lim_{n \to \infty}
  \mathsf{P} \! \left( \frac{S_{n} - n p}{\sqrt{n p (1 - p)}} \leq z \right).
$$
This is precisely the statement of the de Moivre-Laplace Theorem.
