Asymptotic standard normal distribution I need to solve the following exercise.

Assume that $X_\lambda$ is Poisson distributed with mean $\lambda$ . Show that $Y(\lambda) = \frac{X_\lambda - \lambda}{\sqrt{\lambda}}$ is asymptotic standard normal distributed for $\lambda \to \infty$.

Any help would be really appreciated.
 A: Not quite sure if the one on Wiki is right or if I understand it, but anyway I'm basing my answer on CLT in Larsen and Marx (an elementary probability book):


Anyway, this answer might be weird or very wrong or stupid but...
Consider iid $X_i$ ~ $Poi(\lambda/n)$.
Then $X := \sum_{i=1}^{n} X_i$ ~ $Poi(\lambda)$
Then
$$(\frac{1}{n} \sum_{i=1}^{n} X_i - \mu) / (\sigma/\sqrt{n})$$
$$= (\frac{1}{n} \sum_{i=1}^{n} X_i - \lambda/n) / (\sqrt{\lambda}/n)$$
$$= (\sum_{i=1}^{n} X_i - \lambda) / (\sqrt{\lambda})$$
$$= Y(\lambda)$$

Now I'm thinking we have to come up w/ some kind of relationship between n and $\lambda$ s.t. as $n \to \infty$, $\lambda \to \infty$

Guess 1:
$$= (\sum_{i=1}^{(\frac{\lambda}{E[X_i]})} X_i - \lambda) / (\sqrt{\lambda})$$
I was thinking as $\lambda \to \infty \iff n \to \infty$ as $\lambda = n E[X_i]$ but then again $E[X_i]$ is a function of both n and $\lambda$

Guess 2:
$P(X_i = k) \ge 0 \to (\frac{\lambda}{n})^k \ge 0$
Not sure how to proceed. I guess $\exists a > 0$ s.t. $\lambda^k = a n^k$ ?
