Assume $X_n$ is distributed $\text{Beta}(1/n, 1/n)$ and $X$ is distributed as $\text{Binom}(1,1/2)$. Show that $X_n$ converges to $X$ in distribution.  I'm having some issue with this question. I understand that if I want to show that if $X_n$ converges to $X$ in distribution then $F(X_n) \rightarrow F(x)$. This is for their cumulative distribution functions. However, I'm a bit stuck on how exactly to go about this.
It's clear to me by inspection that if we fix $\beta = \frac{1-\mu}{\mu} \alpha$ (thereby fixing the mean) and let $\alpha \rightarrow 0$, the Beta distribution approaches a Bernoulli($\mu$) distribution.  For example:  par(mfrow = c(1, 2),     oma = c(0, 0, 1.5, 0)) xx = seq(0, 1, length.out = 1000) mus = c(.2, .7) for (ii in 1:2) {   mu = mus[ii]   matplot(xx, sapply(10^(-1:-5), function(al)      pbeta(xx, al, (1-mu)/mu * al)),     type = 'l', lty = ii,     main = paste('Mean:', mu),     ylab = 'Cumulative Probability', xlab = 'x') } title('Beta Approaches Bernoulli', outer = TRUE)     That is, if $X \sim B(\alpha, \frac{1-\mu}{\mu} \alpha)$, its CDF $F_X(x; \alpha, \mu)$, satisfies  $$ \lim_{\alpha \rightarrow 0} F_X(x; \alpha, \mu) = 1 - \mu = F_Y(x) \quad \forall x \in (0, 1)$$  Where $Y \sim$ Bernoulli($\mu$), and the convergence is non-uniform. I tried to do a more formal proof of this but couldn't make any progress, even in simple cases (like $\mu = \frac12$). The  makes reference to the $\mu = \frac12$ case several times without proof. Perhaps I'm missing something easy about doing the integral, or some fact about the Beta/incomplete Beta function. Any help would be appreciated!
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