The normal approximation of Poisson distribution (I've read the related questions here but found no satisfying answer, as I would prefer a rigorous proof for this because this is a homework problem)
Prove: If $X_\alpha$ follows the Poisson distribution $\pi(\alpha)$, then
$$\lim_{\alpha\rightarrow\infty}P\{\frac{X_\alpha-\alpha}{\sqrt{\alpha}} \leq u \} = \Phi(u)$$
where $\Phi(u)$ is the cdf of normal distribution $N(0,1)$
Hint: use the Laplace transform $E(e^{-\lambda(X_\alpha-\alpha)/\sqrt{\alpha}})$, show that as $\alpha\rightarrow\infty$ it converges to $e^{\lambda^2/2}$
I did the transform but failed to sum the series(which is essentially doing nothing)

Here's what I got:
$$g(\lambda)=\sum_{n=0}^{\infty} \frac{e^{-\alpha}}{n!}\alpha^n e^{-\frac{\lambda(n-\alpha)}{\sqrt{\alpha}}}$$
and $\lim_{\alpha\rightarrow\infty} g(\lambda)=e^{-\lambda^2}$ is what I'm trying to arrive at. I tried L'Hospital only to find that the result is identical to the original ratio.
 A: Let $X_{\alpha}$  Poisson $\pi(\alpha)$, for $\alpha = 1, 2, \ldots$ 
The probability mass function of  $X_{\alpha}$ is 
$$f_{X_{\alpha}}(x)=\frac{{\alpha}^x\operatorname{e}^{-x}}{x!} \qquad \alpha = 1, 2, \ldots$$
The moment generating function of $X_{\alpha}$ is
$$
M_{X_{\alpha}}=\Bbb{E}\left(\operatorname{e}^{tX_{\alpha}}\right)=\operatorname{e}^{\alpha(\operatorname{e}^{t}-1)}\qquad t\in(-1,1)
$$
Now consider a “standardized” Poisson random variable $Z_{\alpha}=\frac{X_{\alpha}-{\alpha}}{\sqrt{\alpha}}$
which has limiting moment generating function
$$
\begin{align}
\lim_{\alpha\to\infty}M_{Z_{\alpha}}&=
\lim_{\alpha\to\infty}\Bbb{E}\left(\exp{(tZ_{\alpha})}\right)\\
&=\lim_{\alpha\to\infty}\Bbb{E}\left(\exp{\left(t\frac{X_{\alpha}-{\alpha}}{\sqrt{\alpha}}\right)}\right)\\
&=\lim_{\alpha\to\infty}\exp(-t\sqrt\alpha)\Bbb{E}\left(\exp{\left(\frac{tX_{\alpha}}{\sqrt{\alpha}}\right)}\right)\\
&=\lim_{\alpha\to\infty}\exp(-t\sqrt\alpha)\exp\left(\alpha(\operatorname{e}^{t/\sqrt\alpha}-1)\right)\\
&=\lim_{\alpha\to\infty}\exp\left(-t\sqrt\alpha +\alpha\left[t\alpha^{-1/2}+\frac{t^2\alpha^{-1}}{2}+\frac{t^3\alpha^{-3/2}}{6}+\cdots\right]\right)\\
&=\lim_{\alpha\to\infty}\exp\left(\frac{t^2}{2}+\frac{t^3\alpha^{-3/2}}{6}+\cdots\right)\\
&=\exp\left(\frac{t^2}{2}\right)
\end{align}
$$
by using the moment generating function of a Poisson random variable and expanding the
exponential function as a Taylor series. This can be recognized as the moment generating function of a standard normal random variable. This implies that the associated unstandardized random variable $X_{\alpha}$ has a limiting distribution that is normal with mean $\alpha$ and variance $\alpha$.
