Showing that $\frac{X_n -n}{\sqrt{X_n}}\to N(0,1)$ in distribution where $X_n\sim \operatorname{Gamma}(n,1)$ We're given that $X_n\sim \operatorname{Gamma}(n,1)$ and we want to show that $Y_n \rightarrow N(0,1)$ in distribution where $Y_n=\frac{X_n -n}{\sqrt{X_n}}$
A little stuck on this one.
Using the Weak Law of Large numbers, we know that $X_n$ converges in distribution to $E(X_n) = n$.
We can then use Chebyshev's Inequality to show that $\bar{X_n}$ converges in distribution to $1$
$$P(|X_n - n| ≥ c) = P(|\bar{X_n} - 1| ≥ c) ≤ \frac{\operatorname{Var}(\bar{X_n})}{c^2} = \frac{\operatorname{Var}(X_n)}{n^2c^2} = \frac{n}{n^2c^2} = \frac{1}{nc^2} $$
Since this goes to $0$ as $n \rightarrow \infty$, $X_n$ converges to $1$ in distribution. We can continue and say $\sqrt{X_n}$ converges to $1$ in distribution since $1$ is continuous there.
From here I'm stuck, what will take me to the final answer, Since the denominator converges to $1$, we just need to show the numerator converges to $N(0,1)$ by Slutsky's Theorem, I'm just a little lost on how to get there.
 A: Let $(W_n)_n$ be sequence of random variables independent and identically distributed following exponential distribution $\mathcal{E}(1),$ let $S_n=\sum_{k=1}^nW_k.$
Notice that $Y_n=\sqrt{\frac{n}{X_n}}\frac{X_n-n}{\sqrt{n}},$ and that $S_n$ follows gamma distribution $Gamma(n,1),$ using the central limit theorem and the strong law of large number, we get that $\sqrt{\frac{n}{S_n}}$ converges in distribution to $1$ and that $\frac{S_n-n}{\sqrt{n}}$ converges in distributiont to $N(0,1),$ conclude using Slutsky's lemma. 
A: A nice property of the Gamma distribution is that if $X \sim \Gamma(r_1, \lambda)$ and $Y \sim \Gamma(r_2, \lambda)$, then $X + Y = \Gamma(r_1 + r_2, \lambda)$. We can thus conclude by induction that if we have an infinite sequence of independent random variables $(W_k)_{k \in \mathbb{N}}$ where $W_k \sim \Gamma(1, 1)$ for all $k \in \mathbb{N}$, then
$$\left(\sum_{k=1}^n W_k \right) \sim \Gamma(n, 1).$$
The central limit theorem tells us that if $(W_k)_{k \in \mathbb{N}}$ is a sequence of independent and identically distributed random variables, then
$$\frac{\left(\sum_{k=1}^n W_k\right) - n \mu}{\sigma \sqrt{n}} \rightarrow N(0, 1),$$
where $\mu = \mathbb{E}(W_k)$ and $\sigma^2 = \text{Variance}(W_k)$. The gamma distribution $\Gamma(r, \lambda)$ has expectation $\frac{r}{\lambda}$ and variance $\frac{r}{\lambda^2}$. In this case, $r = 1 = \lambda$, so the expectation and variance are both 1. Therefore, we have that
$$\frac{\left(\sum_{k=1}^n W_k\right) - n}{\sqrt{n}} \rightarrow N(0, 1),$$
and since we saw before that
$$\left(\sum_{k=1}^n W_k \right) \sim \Gamma(n, 1),$$
we get the desired result. Hope this helps! Let me know if anything is unclear.
