The proof of an elementary equality For all $t\geq0$, $x>0$, we have the following important equality:
\begin{equation}
\lim_{\lambda\rightarrow\infty}e^{-\lambda t}\sum_{k\leq\lambda x}\frac{(\lambda t)^k}{k!}=\chi_{[0,x)}(t)+\frac{1}{2}\chi_{\{x\}}(t),
\end{equation}
where $\chi$ is the characteristic function. Actually, I can prove this equality with the help of the Poisson distribution, but I prefer to prove this equality by a basic analysis method. I've struggled with this problem for a few days, still have no idea.
I'll appreciate it for any hints! Thanks for any help!
 A: Partial answer: Suppose $t\in[0,x)$, then we want to show the limit on the left hand side of the equation is $1$. Writing it out, $$e^{-\lambda t}\sum_{k\leq\lambda x}\frac{(\lambda t)^k}{k!} = e^{-\lambda t}\left(e^{\lambda t}-\sum_{k >\lambda x}\frac{(\lambda t)^k}{k!}\right)=1-e^{-\lambda t}\sum_{k >\lambda x}\frac{(\lambda t)^k}{k!},$$
so we just need to show the second term goes to zero as $\lambda\to\infty$. Let $k_0=\lfloor\lambda x\rfloor+1$ be the first term that occurs in the sum. Then we have the bound $$\begin{split}e^{-\lambda t}\sum_{k >\lambda x}\frac{(\lambda t)^k}{k!}&<e^{-\lambda t}\frac{(\lambda t)^{k_0}}{k_0!}\left(1+\frac{\lambda t}{k_0}+\frac{(\lambda t)^2}{k_0^2}+\cdots\right)\\&=e^{-\lambda t}\frac{(\lambda t)^{k_0}}{k_0!}\frac{1}{1-\lambda t/k_0}\\&\sim e^{-\lambda t}\frac{(\lambda t)^{\lambda x}}{(\lambda x)!}\frac{1}{1-t/x}\end{split}$$
where we can treat $t,x$ as constant in this confusing expression. From here, I think using Stirling's approximation for $(\lambda x)!$ should finish the job, and the other cases $t=x$ and $t>x$ should be similar (though I haven't written out the details).
A: A similar problem appeared here Using the probabilistic methods described there we may proceed as follows:
Let $Y_\mu$ be a Poisson random variable with parameter $\mu$. Its' characteristic function is given by
$$
\phi_{X_\mu}(t)=e^{\mu(e^{it}-1)}$$
Consider $Z_\mu:=\frac{Y_\mu-\mu}{\sqrt{\mu}}$. THe Characteristic function of $Z_\mu$ is
$$\phi_{Z_\mu}(t)=\exp\Big(\mu\Big(e^{\tfrac{ti}{\sqrt{\mu}}}-1-\tfrac{ti}{\sqrt{\mu}}\Big)\Big)\xrightarrow{\mu\rightarrow\infty}e^{-t^2/2}$$
That is, $Z_\mu$ converges in law to the standard normal distribution.
Now that we gave this, notice that with $\mu=\lambda t$,
$$A_\mu:=e^{-\lambda t}\sum_{k\leq \lambda x}\frac{(\lambda t)^k}{k!}=\mathbb{P}\big[X_\mu\leq \lfloor\mu\tfrac{x}{t}\rfloor\Big]=\mathbb{P}\Big[Z_\mu\leq \frac{\lfloor \mu\alpha\rfloor -\mu}{\sqrt{\mu}}\Big]$$
where $\alpha=x/t$.

*

*When $t=x$, $\alpha=1$ and $\frac{\lfloor \mu\alpha\rfloor -\mu}{\sqrt{\mu}}\approx0$ and so, $A_\mu\xrightarrow{\mu\rightarrow\infty}\Phi(0)=\frac12$

*When $t<x$, $\alpha<1$ and  $\frac{\lfloor \mu\alpha\rfloor -\mu}{\sqrt{\mu}}\approx (\alpha-1)\sqrt{\mu}\xrightarrow{\mu\rightarrow\infty}-\infty$ and so, $A_\mu\xrightarrow{\mu\rightarrow\infty}\Phi(-\infty)=0$

*When $t>x$, $\alpha>1$ and  $\frac{\lfloor \mu\alpha\rfloor -\mu}{\sqrt{\mu}}\approx (\alpha-1)\sqrt{\mu}\xrightarrow{\mu\rightarrow\infty}\infty$ and so, $A_\mu\xrightarrow{\mu\rightarrow\infty}\Phi(\infty)=1$
There are a few details to consider about the right way to pass to the limits in $\mathbb{P}\Big[Z_\mu\leq  \frac{\lfloor \mu\alpha\rfloor -\mu}{\sqrt{\mu}}\Big]$ since $\frac{\lfloor \mu\alpha\rfloor -\mu}{\sqrt{\mu}}$ is not fixed, but the details are not complicated (Slutsky's theorem for example).
