Convergence in distribution/Distribution of X For each $n = 1, 2, ....$, suppose that $X_n$ is a discrete random variable with range $\{1/n, 2/n, ..., 1\}$ and 
$\hspace{15mm}\mathrm{Pr}(X_n = j/n) = \frac{2j}{n(n+1)}$, $j = 1,...,n$.
Does $X_1, X_2, ...,$ converge in distribution to some random variable? 
If it does, what is the distribution of $X$?
Attempt:
As $n \rightarrow\infty$, then
$\hspace{15mm}\mathrm{Pr}(X = 0) = 0$, $j = 1,...,n$.
I believe this is incorrect (I'm horrible at proofs of convergence in distribution and convergence), so any help would be much appreciated. 

Fix some $o\leq x\leq 1$. For any $n$, define $k(n)$ by $\frac{k(n)}{n}\leq x<\frac{k(n)+1}{n}$. We get
$F_n(x)=P(X_n\leq x)=\sum_{j=1}^{k(n)}\frac{2j}{n(n+1)}=\frac{2}{n(n+1)}\sum_{j=1}^{k(n)}j=\frac{2}{n(n+1)}\cdot\frac{k(n)(k(n)+1)}{2}=\frac{k(n)(k(n)+1)}{n(n+1)}$.
Calculate the limit as $n\to\infty$:
$$\lim_{n\to\infty}F_n(x)=\lim_{n\to\infty}\frac{k(n)(k(n)+1)}{n(n+1)}=\lim_{n\to\infty}\left(\frac{k(n)}{n}\cdot\frac{k(n)+1}{n}\cdot\frac{n}{n+1}\right).$$
The left term and the middle one converge to $x$, while the right one converges to $1$, hence
$$\lim_{n\to\infty}F_n(x)=x^2.$$
Define a random variable $X$ by having distribution $F(x)=x^2$ for $0\leq x\leq1$, $F(x)=0$ for negative $x$, and $F(x)=1$ for $x\ge1$, and by definition $X_n\to X$ in distribution.
