Question: Let $X$ be a binomial random variable with parameters $n,p$. Let $q=1-p$. What is the limiting distribution of $a(X+b)$ for large $n$, where $a=\sqrt{npq}$ and $b=-np$?
My attempt: The moment generating function of $a(X+b)$ is
$\mathcal{M}_{a(X+b)}(t)=(pe^{a(1-p)t}+(1-p)e^{-apt})^n$.
The power series expansion of the bracketed term is $1 + \dfrac{1}{2} t^2 (a^2 pq) + \dfrac{1}{6} a^3 p (2 p^2 - 3 p + 1) t^3 +... = 1 + \dfrac{1}{2} n(pq)^2t^2+...$
Here, I get stuck, as I was hoping for the expansion to be $1+\dfrac{t^2}{2n}$ which would give the limiting distribution as $\mathcal{N}(0,1)$. How can I proceed?
In fact $a=\dfrac{1}{\sqrt{npq}}$ because the "normalizing" formula is
$$Y=\dfrac{X-m}{\sigma}$$
with mean $m=np$ and standard deviation $\sigma=\sqrt{npq}$ for a Bin(n,p) distribution.