I quote Jacod-Protter herebelow.
Let $\left(X_n\right)_{n\geq1}$ be a sequence of Poisson random variables with parameter $\lambda_n=n$.
($\ldots$)
We have \begin{equation*} \begin{split} E\big\{e^{iuZ_n}\big\}&=E\Bigg\{e^{iu\left(\frac{1}{\sqrt{n}}\left(X_n-n\right)\right)}\Bigg\}\\ &=e^{-iu\sqrt{n}}E\Big\{e^{i\frac{u}{\sqrt{n}}X_n}\Big\}\\&=e^{-iu\sqrt{n}}e^{n\left(e^{iu/\sqrt{n}-1}-1\right)} \end{split} \end{equation*} where the last equality follows from the fact that if $X\sim\text{Poisson}\left(\lambda\right)$, $\varphi_X(u)=E\{iuX\}=e^{\lambda\left(e^{iu}-1 \right)}$.
Continuing and using a Taylor expansion for $e^z$, we have the above equals \begin{equation*} \begin{split} &=e^{-iu\sqrt{n}}e^{n\left(i\frac{u}{\sqrt{n}}-\frac{u^2}{2n}-\frac{iu^3}{6n^{3/2}}+ \hspace{0.2cm}\cdots\right)}\\&=e^{-iu\sqrt{n}+iu\sqrt{n}}e^{-u^2/2}e^{-\frac{h(u, n)}{\sqrt{n}}}\\&=e^{-u^2/2}e^{-\frac{h(u, n)}{\sqrt{n}}} \end{split} \end{equation*} where $h(u,n)$ stays bounded in $n$ for each $u$ and hence $\lim\limits_{n\to\infty}\frac{h(u,n)}{\sqrt{n}}=0$
I got three doubts referring to part in bold above:
1. Why is one allowed to precisely say that $-\frac{iu^3}{6n^{3/2}}+ \hspace{0.2cm}\cdots=\frac{h(u,n)}{\sqrt{n}}$?;
2. What does it mean that "$h(u,n)$ stays bounded in $n$ for each $u$"? And how can I show this?;
3. $\lim\limits_{n\to\infty}\frac{h(u,n)}{\sqrt{n}}=0$ since $\sqrt{n}$ goes to $+\infty$ faster than $h(u,n)$, right? If so, I would argue that $h(u,n)$ is of an infinitesimal order which is lower than that of $\sqrt{n}$, but I cannot see why it is so (as it is clear from point $2.$ right above)
Okey, let $X_n \sim Poiss(n)$ and let $Z_n = \sqrt{n}(Z_n- n)$. We want to calculate $\varphi_n(u) = \mathbb E[\exp(iuZ_n)]$. We get:
$\varphi_n(u) = \exp(-iu\sqrt{n})\mathbb E[\exp(iu \frac{X_n}{\sqrt{n}})] = \exp(-iu\sqrt{n})e^{n(\exp(iu\frac{1}{\sqrt{n}})-1)}$.
Now, $\exp(z) = \sum_{k=0}^\infty \frac{z^k}{k!}$, so that we can write: $\exp( iu\frac{1}{\sqrt{n}})= \sum_{k=0}^\infty \frac{1}{k! (\sqrt{n})^k}i^ku^k = 1 + iu\frac{1}{\sqrt{n}} -\frac{u^2}{2n}+ \sum_{k=3}^\infty \frac{1}{k! (\sqrt{n})^k}i^ku^k$, which gives:
$$\varphi_n(u) = \exp(-\frac{u^2}{2})\exp(\sum_{k=3}^\infty \frac{1}{k! (\sqrt{n})^k}i^ku^k)$$
Now, what's done in your case, was just saying that $\sum_{k=3}^\infty \frac{1}{k! (\sqrt{n})^k}i^ku^k = \frac{h(n,u)}{\sqrt{n}} $
$\bf {Question \ 1}$: They just defined function $h$ in that way, there is nothing more.
$\bf {Question \ 2}$: It means that for fixed $u \in \mathbb R$, sequence $(h(n,u))_{n \in \mathbb N}$ is bounded, that is $\forall _{u \in \mathbb R} \exists_{M_u \in \mathbb R_+}$ such that $\forall_{n \in \mathbb N}: |h(n,u)| \le M_u$.
And why is that? Note that $|h(n,u)| = |\sum_{k=3}^\infty \frac{1}{k! (\sqrt{n})^{k-1}}i^ku^k | \le \sum_{k=3}^\infty \frac{|u|^k}{k! \sqrt{n}^{k-1}} \le \sum_{k=3}^\infty \frac{|u|^k}{k!} \le \exp(|u|)$, so for fixed $u$, sequence $h(n,u)$ is bounded.
$\bf {Question \ 3}$: Using 2), we get for any fixed $u \in \mathbb R$, that $0 \le \frac{|h(n,u)|}{\sqrt{n}} \le \frac{\exp(|u|)}{\sqrt{n}} \to 0$, so for any fixed $u \in \mathbb R$, we get that $\frac{h(n,u)}{\sqrt{n}}$ tends to $0$ as $n \to \infty$, so for any fixed $u \in \mathbb R$, we get $\varphi_n(u) \to \exp(-\frac{u^2}{2})$ and now apply Levy continuous Theorem.