Show the Binomial distribution approaches a Normal distribution (using characteristic function) Let $X_n = Bin(n,p)$. We fix $p$, and we want to show that as $n \to \infty$, $\frac{X_n-np}{\sqrt{np(1-p)}}$ converges to $N(0,1)$ in distribution. And I want to show this using characteristic functions, not using the CLT.
The characteristic function of $X_n$ is $(pe^{it} + (1-p))^n$, so the CF of $\frac{X_n-np}{\sqrt{np(1-p)}}$  is $$\left(pe^{it \frac{1}{\sqrt{np(1-p)}}}+(1-p)\right)^n\ e^{\frac{-it np}{\sqrt{np(1-p}}}.$$
I'm having trouble finding the limit as $n \to \infty$, as I don't think $e^{\frac{-it np}{\sqrt{np(1-p}}}$ has a limit as $n \to \infty$, and I've tried putting $(pe^{it \frac{1}{\sqrt{np(1-p)}}}+(1-p))^n$ into the form $\left(1+\frac{a_n}{n}\right)^n$ with $a_n \to a$, so we can find the limit to be $e^a$, but that has been to no avail.
Thanks.

Apply the expension of exponential and logrithm. The CF of $(X_n - np) / \sqrt{np(1-p)}$ is
$$
\begin{align}
f(t) &= \left(pe^{it\frac{1}{\sqrt{np(1-p)}}} -p + 1 \right)^ne^{-it\frac{np}{\sqrt{np(1-p)}}} \\
&= \exp\{n\left(\log\left(pe^{it\frac{1}{\sqrt{np(1-p)}}} -p + 1\right)\right) -it\frac{np}{\sqrt{np(1-p)}}\} \\
&\sim\exp\{n\left(\log\left(p{it\frac{1}{\sqrt{np(1-p)}}} - \frac12p{t^2\frac{1}{{np(1-p)}}}+o(n^{-3/2})+ 1\right)\right) -it\frac{np}{\sqrt{np(1-p)}}\} \\
&\sim \exp\{ n\left( pit\frac1{\sqrt{np(1-p)}}- \frac12p{t^2\frac{1}{{np(1-p)}}}+\frac12\frac{p^2t^2}{np(1-p)} + o(n^{-3/2})\right) -it\frac{np}{\sqrt{np(1-p)}}\} \\
&\to \exp\{-\frac12t^2\}
\end{align}
$$
