Recall that a random variable $X$ follows the Poisson distribution $\mathcal{P}(\lambda)$ if $\mathrm{P}(X = n) = \frac{\lambda^{n}}{n!} \mathrm{e}^{-\lambda}$ for all $n \in \mathbb{N}$. Let $\left(X_{n}\right)_{n \geqslant 1}$ be a sequence of mutually independent random variables with distribution $\mathcal{P}(\lambda)$, $S_{n} = X_{1} + \cdots + X_{n}$, and $T_{n} = \frac{S_{n} - n\lambda}{\sqrt{n\lambda}}$. Determine an equivalent, when $n \rightarrow +\infty$, of $$A_{n} = \sum_{k=0}^{\lfloor n\lambda \rfloor} \frac{(n\lambda)^{k}}{k!} \quad \text{and} \quad B_{n} = \sum_{k=\lfloor n\lambda \rfloor + 1}^{+\infty} \frac{(n\lambda)^{k}}{k!}$$ where $\lfloor t \rfloor$ denotes the integer part of the real number $t$. We will interpret $\mathrm{e}^{-n\lambda} A_{n}$ as the probability of an event related to $S_{n}$ and thus to $T_{n}$.
Recall that a random variable $X$ follows the Poisson distribution $\mathcal{P}(\lambda)$ if $\mathrm{P}(X = n) = \frac{\lambda^{n}}{n!} \mathrm{e}^{-\lambda}$ for all $n \in \mathbb{N}$. Let $\left(X_{n}\right)_{n \geqslant 1}$ be a sequence of mutually independent random variables with distribution $\mathcal{P}(\lambda)$, $S_{n} = X_{1} + \cdots + X_{n}$, and $T_{n} = \frac{S_{n} - n\lambda}{\sqrt{n\lambda}}$.
Determine an equivalent, when $n \rightarrow +\infty$, of
$$A_{n} = \sum_{k=0}^{\lfloor n\lambda \rfloor} \frac{(n\lambda)^{k}}{k!} \quad \text{and} \quad B_{n} = \sum_{k=\lfloor n\lambda \rfloor + 1}^{+\infty} \frac{(n\lambda)^{k}}{k!}$$
where $\lfloor t \rfloor$ denotes the integer part of the real number $t$.
We will interpret $\mathrm{e}^{-n\lambda} A_{n}$ as the probability of an event related to $S_{n}$ and thus to $T_{n}$.