I'm learning convergence and struggling with this problem.
Let there be a sequence of r.v.'s $\{ X_{n}\}_{n\geq1}$ with the following density
$$f_{X_n}=\frac{nx^{n-1}}{\theta^n}, 0 < x< \theta$$
Show that $X_n \xrightarrow[]{D}\theta$.
I know that $P(X_{n} \leq a)\xrightarrow[]{}P(X\leq a)$ when $n \xrightarrow[]{}\infty$ is a sufficient condition to the convergence above, so we need to integrate it and apply the limit:
$$\int_0^a \frac{nx^{n-1}}{\theta^n} \, dx = \left.\frac{nx^n}{n\theta^n}\right\rvert_0^a = \frac{a^{n}}{\theta^n}$$
How should i proceed from here? I don't see how applying the limit here will get me to $\theta$. Thanks in advance for the help.
Let $F_n$ be the distribution function of $X_n$. Then $X_n$ converges in distribution to a distribution $F$ if and only if $F_n(x)\stackrel{n\to\infty}\longrightarrow F(x)$ for all $x$ at which $F$ is continuous. Here $$F_n(x) = (x/\theta)^n\cdot\mathsf 1_{(0,\theta)}(x) + \mathsf 1_{[\theta,\infty)}(x)$$ and since $0<x/\theta<1$, we have $$F_n(x)\stackrel{n\to\infty}\longrightarrow \mathsf 1_{[\theta,\infty)}(x)=:F(x).$$ As $F(\theta)=1$ and $F(x)=0$ for $x<\theta$, it follows that $X_n$ converges in distribution to $\theta$.