Showing that the symmetric Beta distribution is normal by Slutsky's Theorem? Suppose that $B_n \sim B(n,n)$, then I would like to show that as $n \to \infty$, $2\sqrt{2n}(B_n-\frac{1}{2}) \to N(0,1)$. 
I know that the Beta distribution can be written as the following Gamma form:
$$
B_n = \frac{G_n}{G_n + G'_n}
$$
Where $G_n$ is a Gamma distribution with shape parameter $n$ and scale parameter $1$, and $G'_n$ is a Gamma distribution with shape parameter $n$ and scale parameter $1$ independent of $G_n$. 
Now, the Gamma in turn can be rewritten as the sum of Exponential distribution with scale $1$. 
So, 
$$
G_n = \sum_{i=1}^n X_i 
$$
where $X_i \sim Expo(1)$ which are iid of each other and that 
$$
G'_n = \sum_{i=1}^n X'_i 
$$
where $X'_i \sim Expo(1)$ are iid of each other as well. 
Hence,
$$
B_n = \frac{\sum_{i=1}^n X_i }{\sum_{i=1}^n X_i  + \sum_{i=1}^n X'_i } = \frac{\frac{1}{n}\sum_{i=1}^n X_i}{\frac{1}{n}\sum_{i=1}^n X_i  + \frac{1}{n}\sum_{i=1}^n X'_i } 
$$
Now, by the Central Limit Theorem and Delta Method, the numerator has the distribution representation:
$$
\sqrt{n}\left(\frac{1}{n}\sum_{i=1}^n X_i - 1 \right) \overset{D}\to N\left(0,1\right)
$$
and that the bottom, by Law of Large Numbers, has that:
$$
\frac{1}{n}\sum_{i=1}^n X_i \overset{P}\to 1 \ \ \ \text{and} \ \ \ \frac{1}{n}\sum_{i=1}^n X'_i \overset{P}\to 1
$$
Then, 
$$
\frac{1}{n}\sum_{i=1}^n X_i  + \frac{1}{n}\sum_{i=1}^n X'_i  \overset{P}\to 2
$$
But because convergence in probability implies convergence in distribution, we have that 
$$
\frac{1}{n}\sum_{i=1}^n X_i  + \frac{1}{n}\sum_{i=1}^n X'_i  \overset{D}\to 2
$$
Now, by Slutsky's theorem,
we have that 
$$
\sqrt{4n}\left(B_n - \frac{1}{2}\right) = N\left(0,1\right)
$$
HOWEVER, it is my understanding that the symmetric Beta distribution converges towards $\sqrt{8n}\left(B_n - \frac{1}{2}\right) = N\left(0,1\right)$.
I am off by a factor of $\sqrt{2}$, what did I do wrong here?
 A: The issue in your solution is that
$$ \sqrt{n}\bigg( \frac{1}{n}\sum_{k=1}^{n} X_k - \frac{1}{2} \bigg) \not\Rightarrow \mathcal{N}(0, 1). $$
A correct application of CLT will tell you that
$$ \sqrt{n}\bigg( \frac{1}{n}\sum_{k=1}^{n} X_k - \color{red}{\boxed{1}} \bigg) \Rightarrow \mathcal{N}(0, 1) $$
Now let us utilize CLT and Slutsky's theorem to prove the convergence. Indeed, using your realization of $B_n$, we can write
$$
\sqrt{8n} \big( B_n - \tfrac{1}{2} \big)
= \sqrt{2n} \cdot \frac{G_n - G'_n}{G_n + G'_n}
= \frac{2}{\frac{1}{n}(G_n + G'_n)} \cdot \frac{1}{\sqrt{2n}} \sum_{k=1}^{n} (X_k - X'_k).
$$
On the one hand, WLLN tells you that
$$ \frac{1}{n}(G_n + G'_n) \stackrel{p}{\to} 2.$$
On the other hand, CLT tells you that
$$ \frac{1}{\sqrt{2n}} \sum_{k=1}^{n} (X_k - X'_k) \Rightarrow \mathcal{N}(0, 1). $$
Therefore by Slutsky's theorem you can conclude that $\sqrt{8n}(B_n - \frac{1}{2}) \Rightarrow \mathcal{N}(0, 1)$.

Addendum. The issue in your original argument was that both the numerator and the denominator of $B_n$ contribute to the fluctuation and that you ignored the contribution from the denominator. To make things work (at least heuristically), let us write
$$ \frac{1}{n}\sum_{k=1}^{n} X_k = 1 + \Delta \bar{X}
\qquad \text{and} \qquad
\frac{1}{n}\sum_{k=1}^{n} X'_k = 1 + \Delta \bar{X}'.$$
We know that $\Delta \bar{X}$ and $\Delta \bar{X}'$ are small quantities, so we can use Taylor expansion to approximate $B_n$ as follows:
$$ B_n = \frac{1 + \Delta \bar{X}}{2 + \Delta \bar{X} + \Delta \bar{X}'} = \frac{1}{2} + \frac{\Delta \bar{X}}{4} - \frac{\Delta \bar{X}}{4} + \text{[higher order terms]}$$
Since both $\Delta \bar{X}$ and $\Delta \bar{X}'$ are independent and have asymptotic distribution $\mathcal{N}(0, \frac{1}{n})$, this gives
$$ B_n \approx \frac{1}{2} + \frac{1}{4}\mathcal{N}\bigg(0, \frac{2}{n}\bigg) = \mathcal{N}\bigg(\frac{1}{2}, \frac{1}{8n} \bigg). $$
