Convergence of independent $\mathcal U {(n,n^2)}$ random variables? What does this sequence of random variable's distribution converge to? The random variables are given as follows
$$Y_n=\frac{X_n-n}{n^2}, \quad n=1,2,3, \dots$$ and $X_n\sim\mathcal{U(n,n^2)}$- a sequence of independent variables. The answer is given by: 
$$P(Y_n<y)=P\left(\frac{X_n-n}{n^2}<y\right)=\begin{cases}0, &y\leq 0 \\[0.2cm] \dfrac{ny}{n-1}, &0<y \leq 1- \frac{1}{n} \\[0.2cm] 1, &y>1-\frac{1}{n} \end{cases}:n\to \infty =\mathcal U(0,1)$$ 
The interval that I do not understand is $$0<y \leq 1- \frac{1}{n}$$
Can someone explain this?

You're asking two different questions, one in your title about what $Y_n$ converges to and one about the behavior of $Y_n<y$ for $0<y\leq 1-1/n$. For the latter,
$$P(Y_n<y)=P(X_n< n^2y+n),$$
and since $X_n$ is uniform in $[n,n^2]$, you'll need to figure out when $n^2y+n=n^2$, achieving the upper bound of $X_n$, which gives $y=1-1/n$. So for $y\geq 1-1/n$, $P(Y_n<y)=1$ and otherwise:
$$P(X_n<n^2y+n)=\frac{(n^2y+n)-n}{n^2-n}=\frac{ny}{n-1}$$
