Central Limit Theorem Proof using Characteristic Functions I have my proof of the Central Limit Theorem using characteristic functions in which the main steps are: 
Set up: Let $X_1, X_2, \dots, X_n$ be a sequence of iid random variables with mean $\mu$ and variance $\sigma^2$. Let $Y_i=\frac{X_i-\mu}{\sigma}$ and $S_n=\sum_{i=1}^{n}\frac{1}{\sqrt{n}}Y_i$.
1) Expand the characteristic function $\psi_{S_n}(t)=\psi_{Y}\left(\frac{t}{\sqrt{n}}\right)^n$
2)Taylor expand the characteristic function:
$$
\psi_{S_n}=(1+\frac{it}{\sqrt{n}}\mathbb{E}(Y)-\frac{t^2}{2n}\mathbb{E}(Y^2)+\dots)^n
$$
Which we can then truncate with a bounded function $H(t)$:
$$
\psi_{S_n}=(1+\frac{it}{\sqrt{n}}\mathbb{E}(Y)-\frac{t^2}{2n}\mathbb{E}(Y^2)+n^{-\frac{3}{2}}H(t))^n
$$
3) Using $\mathbb{E}(Y)=0$ and $\mathbb{V}(Y)=1$ we get:
$$
\psi_{S_n}=(1+-\frac{t^2}{2n}+n^{-\frac{3}{2}}H(t))^n
$$
Which if we take logs and the limit $n\to\infty$ we can show that, using 
$$
\frac{\log(1+x)}{x}\to1
$$
As $x\to 0$, we get $\log(\psi_{S_n})\to-\frac{t^2}{2}$ and then it exponate each term we recover that $\psi_{S_n}$ tends to the characteristic function of a $N(0,1)$ distribution.
My question is, how do we know that $H(t)$ is bounded and will tend to zero as $n\to\infty$? I thought it had something to do with the moments of a $N(0,1)$ distribution being finite since we have $\mathbb{E}(Y)=0$ and $\mathbb{V}(Y)=1$ but I am not sure this is correct. I have tried to read papers online, but they usually just use little-o notation for this part of the proof, and don't explain how they know that we can essentially disregard this term. Thanks for any help.
 A: I'm not sure that there is such an $H(t)$ in general (if $E(|X|^3)<\infty$ maybe). What is the case is that as long as $X$ has a variance then
$$\phi_X(t)=1+i\mu t-\frac{\sigma^2}{2}t^2+h(t)$$
where $h(0)=0$ (obviously) and $h(t)/t^2\to0$ as $t\to0$.
From now on, I'll take $\mu=0$ and $\sigma=1$ without any real loss of generality.
Fix $t\ne0$. Then
$$\phi_{S_n}(t)=\left(1-\frac{t^2}{2n}+h(n^{-1/2}t)\right)^n.$$
For large enough $n$ then
$$\ln\phi_{S_n}(t)=n\ln\left(1-\frac{t^2}{2n}+h(n^{-1/2}t)\right)
=n\left(-\frac{t^2}{2n}+h(n^{-1/2}t)\right)
+n\Psi\left(-\frac{t^2}{2n}+h(n^{-1/2}t)\right)$$
where we define
$$\Psi(x)=\ln(1+x)-x.$$
Now $nh(n^{-1/2}t)=t^2h(n^{-1/2}t)/(n^{-1/2})^2\to0$
as $n\to\infty$. Also $\Psi(x)/x^2\to-1/2$ as $x\to0$.
If we take $x=-t^2/(2n)+h(n^{-1/2}t)$ we see that $n\Psi(x)\to0$
as $n\to\infty$ (after a little effort).
Now where does $h$ come from? Well,
$$h(t)=E\left(e^{itX}-1-itX+\frac{t^2X^2}2\right)=E(\eta(tX))$$
where
$$\eta(u)=e^{iu}-1-iu+\frac{u^2}2=\sum_{k=3}^\infty\frac{(iu)^k}{k!}.$$
Then $\eta(u)/u^2$ is bounded and tends to $0$ as $u\to0$. Then
$$E(\eta(tX))=t^2E\left(\frac{\eta(tX)}{t^2}\right).$$
Now
$$E\left(\frac{\eta(tX)}{t^2}\right)\to0$$
as $t\to0$ (I think by dominated convergence).
