If $X_n \sim \text{Beta}(n, n)$, show that $[X_n - \text{E}(X_n)]/\sqrt{\text{Var}(X_n)} \stackrel{D}{\longrightarrow} N(0,1)$ Let $X_n \sim \mathbf{B}(n,n)$ (Beta distribution), with pdf
$$
f_n(x) = \frac{1}{\text{B}(n,n)}x^{n-1}(1 - x)^{n-1},~~ x \in (0,1).
$$
Knowing that $\text{E}(X_n) = 1/2$ and that $\text{Var}(X_n) = 1/[4(2n+1)]$, prove that
$$
2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) \stackrel{D}{\longrightarrow} N(0,1).
$$

I thought about doing it by the definition of convergence in distribution, but the cdf of $ 2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}})$ is obscene. I wouldn't know how to calculate the limit $\text{lim}_{n \to \infty} F_{Y_n}(x) $ where $Y_n =  2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}})$.
Then I thought about proving convergence in probability, since converge in probability $\Rightarrow$
convergence in distribution. The problem is that it may not even converge in probability so it would be wasted work.
Edit:
I did some work and this is where I'm at:
Definition. A sequence of random variables $X_1, X_2, ...$, converges in distribution to a random variable X if
$$
\text{lim}_{n \to \infty} F_{X_n}(x) = F_X(x)
$$
So we have to prove that
$$
\text{lim}_{n \to \infty} F_{Y_n}(x) = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-y^2/2}dy
$$
Where $Y_n = 2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) $.
Now,
$$
\begin{align}
P(Y_n \leq x) & = P(2\sqrt{2n + 1}(X_n - \small{\frac{1}{2}}) \leq x) \\
              & = P(X_n - 1/2 \leq \frac{x}{2\sqrt{2n+1}} \\
              & = P(X_n \leq \frac{x}{2\sqrt{2n+1}} + 1/2) \\
              & = F_{X_n} \Bigl( \frac{x}{2\sqrt{2n+1}} + \frac{1}{2} \Bigr) \\
              & = \frac{1}{B(n,n)}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt
\end{align}
$$
We use Stirling's approximation to $\text{B}(n,n)$:
$$
B(a, b) \approx \sqrt{2\pi} \frac{a^{a - 1/2}b^{b - 1/2}}{(a + b)^{a + b - 1/2}}
$$
So $\text{B}(n, n) \approx \frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}} $, after simplification.
Substituting the Stirling approximation (we do this because it converges asymptotically and we're taking the limit), we obtain
$$
\frac{1}{\frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}}}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt.
$$
So what's left to do is prove that
$$
\text{lim}_{n \to \infty}  \frac{1}{\frac{\sqrt{\pi}}{2^{2n - 1}} \frac{1}{\sqrt{n}}}\int_{0}^{ \frac{x}{2\sqrt{2n+1}} + 1/2 } t^{n-1}(1 - t)^{n-1}dt = \int_{-\infty}^{x} \frac{1}{ \sqrt{2\pi}} e^{-y^2/2}dy.
$$
Edit 2: I asked my professor for guidance on how to finish the last step. All he said was "apply the limit theorem to solve directly".
 A: The answer is in portuguese because I'm a native portuguese speaker.
O último cálculo na demonstração acima é um problema computacional excessivamente difícil. Aqui a ideia é apresentarmos uma demonstração alternativa, que se dá pelos seguintes passos:\
1º: Mostramos que a densidade de $ Y_n $ converge para a densidade de $ Z $, onde $ Z \sim N(0, 1). $\
2º: Invocamos o \textit{Lema de Scheffé} para terminar a demonstração. O Lema de Scheffé é um resultado em Teoria da Medida que, no nosso caso, implica que se $ f_{Y_n}(x) \longrightarrow f_Z(x) $, então $ F_{Y_n}(x) \longrightarrow F_Z(x) $, provando a definição Convergência em Distribuição. Em resumo, temos um trabalho facilitado por causa de um resultado mais forte e sofisticado.\
Muito bem, ao diferenciar as equações (1) e (4), obtemos
$$
f_{Y_n}(x) = f_{X_n}(\frac{x}{2\sqrt{2n+ 1} + 1/2}) \frac{1}{2\sqrt{2n + 1}}.
$$
Agora temos que demonstrar que
$$
\text{lim}_{n \to \infty} f_{Y_n}(x) = \text{lim}_{n \to \infty} f_{X_n}(\frac{x}{2\sqrt{2n+ 1} + 1/2}) \frac{1}{2\sqrt{2n + 1}} = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}.
$$
De fato,
\begin{align*}
f_{X_n}(\frac{x}{2\sqrt{2n+ 1} + 1/2}) \frac{1}{2\sqrt{2n + 1}} & = \frac{1}{2\sqrt{2n+1}} \frac{1}{B(n,n)} (\frac{x}{2\sqrt{2n+1}} + \frac{1}{2})^{n-1} (\frac{1}{2} - \frac{x}{2\sqrt{2n+1}})^{n-1}  \\
            & = \frac{1}{2\sqrt{2n+1}} \frac{1}{B(n,n)}(\frac{1}{4} - \frac{x^2}{4(2n+1)})^{n-1} \\
            & = \frac{1}{2\sqrt{2n+1}} \frac{2^{2n - 1} \sqrt{n}}{\sqrt{\pi}} (\frac{1}{4} -  \frac{x^2}{4(2n+1)})^{n-1} \\
            & = \frac{1}{2\sqrt{2n+1}} \frac{2^{2n - 1} \sqrt{n}}{\sqrt{\pi}} (\frac{1}{4})^{n-1}(1 - \frac{x^2}{2n-1})^{n-1} \\
            & = \frac{1}{\sqrt{\pi}} \sqrt{\frac{n}{2n + 1}}(1 - \frac{x^2}{2n-1})^{n-1}.
\end{align*}\
Aplicando o limite $\text{lim}_{n \to \infty}$ na úlima expressão acima, obtemos
\begin{align*}
\text{lim}_{n \to \infty} \frac{1}{\sqrt{\pi}} \sqrt{\frac{n}{2n + 1}}(1 - \frac{x^2}{2n-1})^{n-1} &=  \frac{1}{\sqrt{\pi}} \frac{1}{\sqrt{2}} \text{lim}_{n \to \infty} (1 + \frac{(-x^2/2)}{n -  \frac{1}{2}})^{n-1} \\
&= \frac{1}{\sqrt{2\pi}}e^{-x/2}.
\end{align*}
Mostramos que $ \text{lim}_{n \to \infty} f_{Y_n}(x) = f_Z(x) $. Agora, pelo \textit{Lema de Schéffe}, temos que $ \text{lim}_{n \to \infty} F_{Y_n}(x) = F_Z(x) $, o que prova a convergência em distribuição desejada e termina a demonstração.
