Convergence in distribution - Proof I was given a problem:

For each $n\in\mathbb N$, let $X_n$ be a random variable with uniform distribution over the set $\{0,\frac{1}{n},\frac{2}{n},\dotsc,\frac{n-1}{n},1\}$. Let $U\stackrel{d}{=}\mathcal U[0,1]$ ($U$ has uniform distribution over $[0,1]$). Show that $X_n\stackrel{d}{\underset{n\to\infty}{\longrightarrow}}U.$


My work:
I need to show that $\displaystyle\lim_{n\to\infty}F_{X_n}=F_U$.
Since 
$$f_U(x)=\frac{1}{1-0}=1\Longrightarrow F_U(x)=\cases{0 &if $x<0$\\\int_0^x\,dt=x&if $x<1$\\ 1 & if $x \geq 1$}.$$
Now we have that
$$\mathbb P\left(X_n=\frac{k}{n}\right)=\frac{1}{n+1},k=0,1,\dotsc,n.$$
This leads to
$$F_{X_n}(x)=\cases{0&if $x<0$\\ \sum\limits_{i=0}^{k+1}\frac{1}{n+1}=\frac{k+1}{n+1} &if $\frac{k-1}{n}<x\leq\frac{k}{n}\leq n$\\1 &if $x\geq 1$}.$$
But how do I take the limit here? Wouldn't it be $0$ if $\frac{k-1}{n}<x\leq\frac{k}{n}\leq n$? I appreciate your help.
 A: You're pretty close. Be careful with the distribution function of $X_n$, though. It should be:
$$F_{X_n}(x)=\begin{cases}
0&\text{if}\ x<0,\\
\frac{k}{n+1}&\text{if}\ \tfrac{k-1}n\leq x<\tfrac{k}n,\ k=1,\ldots,n,\\
1&\text{if}\ x\geq1.
\end{cases}$$
Now we need to do an $\epsilon$-proof. Fix $\epsilon>0$ and let $N\in\mathbb{N}$ be larger than $\epsilon^{-1}$. Note that this implies that, for $n\geq N$,
$$0<\frac1{n+1}\leq\frac1{N+1}<\epsilon.$$
All we need to do now is show that if $n\geq N$, $|x-F_{X_n}(x)|<\epsilon$. This is trivial if $x<0$ or $x\geq1$, so assume $0\leq x<1$, and so in particular $\frac{k-1}n\leq x<\frac{k}n$ for some $k$. So $F_n(x)=\frac{k}{n+1}$ and so
$$x-F_{X_n}(x)=\frac{nx+x-k}{n+1}\in\left[\frac{x-1}{n+1},\frac{x}{n+1}\right)$$
where the last step follows from a rearrangement of $\frac{k-1}n\leq x<\frac{k}n$. Since $|x-1|,|x|<1$ the result follows.
A: The characteristic function of $X_n$ is $$\begin{align*}\phi_{n}(t)&=\sum_{k=0}^{n}\frac{1}{n+1}e^{it\frac{k}{n}}=\frac{1}{n+1}\sum_{k=0}^{n}\left(e^{\frac{it}{n}}\right)^k=\frac{1-\left(e^{\frac{it}{n}}\right)^{n+1}}{(n+1)\left(1-e^{\frac{it}{n}}\right)}=\frac{1-e^{it}e^{\frac{it}{n}}}{(n+1)\left(1-e^{\frac{it}{n}}\right)}\end{align*}$$ with $$\begin{align*}\lim_{n\to \infty}\phi_n(t)&=\lim_{n \to\infty}\frac{1-e^{it}e^{\frac{it}{n}}}{(n+1)\left(1-e^{\frac{it}{n}}\right)}\\&=\lim_{h \to0}\frac{1-e^{it}e^{ith}}{\frac{(1+h)\left(1-e^{ith}\right)}{h}}=\lim_{h \to0}\frac{h\left(1-e^{it}e^{ith}\right)}{(1+h)\left(1-e^{ith}\right)}\overset{\frac{0}{0}}{\underset{\text{L' Hopitals rule}}=}\\\\&=\lim_{h \to0}\frac{\left(1-e^{it}e^{ith}\right)+h\left(-ite^{it}e^{ith}\right)}{(1+h)\left(-ite^{ith}\right)+1-e^{ith}}=\frac{e^{it}-1}{it}=\phi_U(t)\end{align*}$$ where $U\sim \mathcal U[0,1]$. This proves the result since convergence of the characteristic functions implies convergence in distribution (and vice versa) i.e. $$\phi_n(t)\to \phi_U(t) \iff X_n \overset{d}\to U$$ 
