(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.