How can I prove that Xn converges to 0 in distribution?

Question

Xn~U[-1/n,1/n]. Since for convergence in distribution Xn-->X iff Fn(x)-->F(x). First of all, I am trying to get the cumulative and it is Fn(x)=(1+xn)/2. Hence, am I doing something wrong?

Answer 1

Hint: Surely $$F_n(x)=\frac{nx+1}{2}$$ but this is only true for $x \in \left[-\frac{1}{n}, \frac{1}{n}\right]$ which goes to $0$ as $n \to \infty$. In particular $$F_n(x)=\begin{cases}0, &x<-\frac{1}{n}\\\frac{nx+1}{2}, &-\frac{1}{n}\le x \le \frac{1}{n}\\ 1, &\frac{1}{n}<x \end{cases}$$ from which you can see that the formula you have above is only valid for the interval $[-1/n, 1/n]$, which converges to the point $0$ as $n \to \infty$.