Convergence in distribution of the following sequence of random variables $X_n\sim Beta\left(\frac{\alpha}{n},\frac{\beta}{n}\right)$ with $\alpha>0$ and $\beta>0$. Does $X_n$ converges to a distribution? How do I approach to show that this converges to a distribution? 
 A: The MGF of a Beta Distribution is:
$$1+\sum_{k=1}^{\infty}\left(\prod_{r=0}^{k-1} \frac{\alpha+r}{\alpha+\beta+r}\right)\frac{t^k}{k!}$$
As $n\to \infty$, we see that 
$$r,\alpha,\beta>0 \implies\frac{\alpha/n+r}{\alpha/n+\beta/n+r} \to 1$$
For $r=0$ we get:
$$\lim_{n\to \infty} \frac{\alpha/n}{\alpha/n+\beta/n} = \frac{\alpha}{\alpha+\beta} = E[X_1]$$
Putting this together we see that:
$$ \lim_{n\to \infty} \left[1+\sum_{k=1}^{\infty}\left(\prod_{r=0}^{k-1}\frac{\alpha/n+r}{\alpha/n+\beta/n+r}\right)\frac{t^k}{k!}\right] = 1+E[X_1]\sum_{k=1}^{\infty}\frac{t^k}{k!} =$$
$$ (1-E[X_1])+E[X_1]e^t$$
The MGF of a $\text{Bernoulli}(p)$ is:
$$(1-p)+pe^t$$
By comparison, we see that the MGF of the Beta has converged to the MGF of a $\text{Bernoulli}\left(\frac{\alpha}{\alpha+\beta}\right)$
It is a known theorem that if random variables $X$ and $Y$ have the same MGF over $t \in (-a,a),a>0$ then they have the same distribution. Clearly, this holds here, so we can conclude, as @Henry alluded to in the comments, that the limiting distribution is indeed a $\text{Bernoulli}\left(\frac{\alpha}{\alpha+\beta}\right)$
A: It converges in distribution to a Bernoulli variable with parameter $\alpha/(\alpha+\beta)$.

This figure shows the Beta distributions in the case $\alpha=2,\beta=1$ for $n=1,4,16,64,$ and $\infty$.  They settle down to a distribution with a jump of $\beta/(\alpha+\beta)=1/3$ at $0$ and another jump of $\alpha/(\alpha+\beta) = 2/3$ at $1$.  This is a Bernoulli$(2/3)$ distribution.

The following sketches a rigorous demonstration using only elementary techniques. 
The idea is to break the Beta integral into three parts: a portion near zero, a portion near one, and everything in between.  We can easily approximate the integrals at the ends using elementary power integrals.  The middle integral becomes much smaller than the two ends, and so eventually can be neglected.  This is obvious when you look at the density function: as soon as $\alpha/n$ drops below $1$, the density shoots up to infinity at the left end.  Similarly, as soon as $\beta/n$ drops below $1$, the density shoots to infinity at the right end.  This produces two "limbs" of a U-shaped distribution.  The limbs dominate the probability.  All that we need to do is (1) show that their relative areas approach a limiting value and (2) compute that limiting value.
For those who might be new to such arguments, here are some details.
Let $1/2 \ge \epsilon \gt 0$.  ($\epsilon$ is going to determine how close to an end of the interval $[0,1]$ we will be.)  We will derive three numerical relationships associated with the Beta PDF $x^{\alpha/n-1}(1-x)^{\beta/n-1}$.  (Notice the lack of the normalizing constant: the trick is to ignore it by looking at relative probabilities.)
First, let $\gamma$ be any number.  From
$$\log(\epsilon^{\gamma/n}) = \frac{\gamma}{n}\log(\epsilon) \xrightarrow{n\to\infty} 0$$
we conclude that $\epsilon^{\gamma/n}$ can be made as close to $\exp(0)=1$ as we like.
Second, for $0 \le x \le 1-\epsilon$, a similar argument yields
$$|\log((1-x)^{\beta/n-1})| = |1 - \beta/n||\log(1-x)| \le |1 - \beta/n||\epsilon| \xrightarrow{n\to\infty} 0.$$
Consequently, for sufficiently large $n$ the value of $(1-x)^{\beta/n-1}$ can be made as close to $\exp(0)=1$ as we desire.  This will later enable us to ignore the contribution of this term to the left-hand limb of the integral.
Third,
$$\int_0^\epsilon x^{\alpha/n-1} dx = \frac{n}{\alpha}\epsilon^{\alpha/n}.$$
Apply these results to the integral of the product:
$$\frac{n}{\alpha}\epsilon^{\alpha/n}= \int_0^\epsilon x^{\alpha/n-1}(1)dx \approx \int_0^\epsilon x^{\alpha/n-1}(1-x)^{\beta/n-1}dx.$$
There's no need to repeat the work for the other limb: the change of variable $x \to 1-x$ interchanges the roles of $\alpha$ and $\beta$, allowing us immediately to write the equivalent approximation
$$\frac{n}{\beta}\epsilon^{\beta/n} \approx \int_{1-\epsilon}^1 x^{\alpha/n-1}(1-x)^{\beta/n-1}dx.$$
Furthermore, applying the foregoing to the case $\epsilon=1/2$ shows us that for sufficiently large $n$
$$\int_\epsilon^{1/2} x^{\alpha/n-1} (1-x)^{\beta/n - 1} dx \approx \frac{n}{\alpha}\left(\left(\frac{1}{2}\right)^{\alpha/n} - \epsilon^{\alpha/n}\right) \ll \frac{n}{\alpha}\epsilon^{\alpha/n}.$$
Again applying the change of variable and adding that result to the preceding result we obtain
$$\int_{\epsilon}^{1-\epsilon} x^{\alpha/n-1}(1-x)^{\beta/n-1}dx \ll \frac{n}{\alpha}\epsilon^{\alpha/n} + \frac{n}{\beta}\epsilon^{\beta/n}.$$
In other words, for large enough $n$ essentially all the probability of a Beta$(\alpha/n,\beta/n)$ distribution is concentrated in the terminal intervals $[0,\epsilon)$ and $(1-\epsilon, 1]$.
The relative probability of the right hand interval, compared to the total probability, therefore comes arbitrarily close to
$$\Pr((1-\epsilon, 1]) \approx \frac{\frac{n}{\beta}\epsilon^{\beta/n}}{\frac{n}{\beta}\epsilon^{\beta/n} + \frac{n}{\alpha}\epsilon^{\alpha/n}} = \frac{\alpha}{\alpha + \beta\epsilon^{(\alpha-\beta)/n} } \xrightarrow{n\to\infty}\frac{\alpha}{\alpha+\beta}$$
and, similarly, the relative probability of the left hand interval comes arbitrarily close to $\beta/(\alpha+\beta)$, QED.
