Can I ask a Limit problem? I've tried a lot of times, but I can't figure out. here is the question:
How to prove:
When $0<t<1$,
$$ \lim_{\lambda \to +\infty}\sum_{k=0}^{[\lambda t]}\frac{\lambda^{k}}{k!}e^{-\lambda} = 0$$
and when $t>1$,
$$ \lim_{\lambda \to +\infty}\sum_{k=0}^{[\lambda t]}\frac{\lambda^{k}}{k!}e^{-\lambda} = 1$$
Thanks so much for your answer!
 A: Suppose $X\sim\operatorname{Poisson}(\lambda)$. In terms of the Iverson bracket, the two claims you want prove can be summarised as one,$\lim_{\lambda\to\infty}P(X\le|\lambda t|)=[t>1]$. You can easily verify this using $X\approx N(\lambda,\,\lambda)$ for $\lambda\gg1$.
A: Note that, given $ \lambda,t\in\mathbb{R} $, we have :
\begin{aligned}\mathrm{e}^{-\lambda}\sum_{k=0}^{\left\lfloor\lambda t\right\rfloor}{\frac{\lambda^{k}}{k!}}&=\frac{1}{\left\lfloor\lambda t\right\rfloor !}\int_{\lambda}^{+\infty}{x^{\left\lfloor\lambda t\right\rfloor}\mathrm{e}^{-x}\,\mathrm{d}x}=1-\frac{\mathrm{e}^{-\lambda}}{\left\lfloor\lambda t\right\rfloor!}\int_{0}^{\lambda}{\left(\lambda -x\right)^{\left\lfloor\lambda t\right\rfloor}\mathrm{e}^{x}\,\mathrm{d}x}\end{aligned}
From the first equality, we get that if $ t<1 $ : \begin{aligned}\mathrm{e}^{-\lambda}\sum_{k=0}^{\left\lfloor\lambda t\right\rfloor}{\frac{\lambda^{k}}{k!}}&=\frac{\mathrm{e}^{-\lambda}}{\left\lfloor\lambda t\right\rfloor !}\int_{\lambda}^{+\infty}{\left(\lambda +x\right)^{\left\lfloor\lambda t\right\rfloor}\mathrm{e}^{-x}\,\mathrm{d}x}\\ &=\frac{\mathrm{e}^{-\lambda}}{\left\lfloor\lambda t\right\rfloor!}\int_{0}^{+\infty}{\left(1+\frac{x}{\lambda}\right)^{\left\lfloor\lambda t\right\rfloor}\mathrm{e}^{-x}\,\mathrm{d}x}\\ &\leq\frac{\mathrm{e}^{-\lambda}}{\left\lfloor\lambda t\right\rfloor!}\int_{0}^{+\infty}{\mathrm{e}^{\frac{\left\lfloor\lambda t\right\rfloor}{\lambda}x}\mathrm{e}^{-x}\,\mathrm{d}x}\\ &\leq\frac{\mathrm{e}^{-\lambda}}{\left\lfloor\lambda t\right\rfloor!}\int_{0}^{+\infty}{\mathrm{e}^{-x\left(1-t\right)}\,\mathrm{d}x}=\frac{\mathrm{e}^{-\lambda}}{\left(1-t\right)\left\lfloor\lambda t\right\rfloor!}\underset{\lambda\to +\infty}{\longrightarrow}0\end{aligned}
And using the second equality, if $ t>1 $ : \begin{aligned}\mathrm{e}^{-\lambda}\sum_{k=0}^{\left\lfloor\lambda t\right\rfloor}{\frac{\lambda^{k}}{k!}}&=1-\frac{\mathrm{e}^{-\lambda}}{\left\lfloor\lambda t\right\rfloor!}\int_{0}^{\lambda}{\left(\lambda -x\right)^{\left\lfloor\lambda t\right\rfloor}\mathrm{e}^{x}\,\mathrm{d}x}\\ &=1-\frac{\mathrm{e}^{-\lambda}\lambda^{\left\lfloor\lambda t\right\rfloor}}{\left\lfloor\lambda t\right\rfloor!}\int_{0}^{\lambda}{\left(1-\frac{x}{\lambda}\right)^{\left\lfloor\lambda t\right\rfloor}\mathrm{e}^{x}\,\mathrm{d}x}\end{aligned}
And thus \begin{aligned}1-\mathrm{e}^{-\lambda}\sum_{k=0}^{\left\lfloor\lambda t\right\rfloor}{\frac{\lambda^{k}}{k!}}&=\frac{\mathrm{e}^{-\lambda}\lambda^{\left\lfloor\lambda t\right\rfloor}}{\left\lfloor\lambda t\right\rfloor!}\int_{0}^{\lambda}{\left(1-\frac{x}{\lambda}\right)^{\left\lfloor\lambda t\right\rfloor}\mathrm{e}^{x}\,\mathrm{d}x}\\&\leq\frac{\mathrm{e}^{-\lambda}\lambda^{\left\lfloor\lambda t\right\rfloor}}{\left\lfloor\lambda t\right\rfloor!}\int_{0}^{\lambda}{\mathrm{e}^{-\frac{\left\lfloor\lambda t\right\rfloor}{\lambda}x}\mathrm{e}^{x}\,\mathrm{d}x}\\ &\leq\frac{\mathrm{e}^{-\lambda}\lambda^{\left\lfloor\lambda t\right\rfloor}}{\left\lfloor\lambda t\right\rfloor!}\int_{0}^{+\infty}{\mathrm{e}^{-x\left(t-1\right)}\,\mathrm{d}x}=\frac{\mathrm{e}^{-\lambda}\lambda^{\left\lfloor\lambda t\right\rfloor}}{\left(t-1\right)\left\lfloor\lambda t\right\rfloor!}\underset{\lambda\to +\infty}{\longrightarrow}0\end{aligned}
