I am trying to solve the following:
Let $X_1,X_2,...$ be a sequence of random variables with $P(X_n=\frac{k}{n})=\frac{1}{n}, k=0,1,2,...,n$. Find the characteristic function of $X_n$ and show that $X \to U(0,1)$ in distribution.
So far I have $\varphi_{X_n}(t)=E[e^{itX_n}]=\sum_{-\infty}^{\infty}e^{itX_n}\frac{1}{n}$. I think there is something wrong here. I am assuming in order to show convergence in distribution I will need that $\lim_{n\to\infty}\varphi_{X_n}(t)\to\frac{e^{it}-1}{it}$ which implies $X \to U(0,1)$ I'm just stuck on the intermediate steps.
First, $k=1,\dots,n$ (otherwise, the total prob. does not sum up to 1). Then
$$\varphi_{X_n}(t)=\mathbb{E}[e^{\mathrm{i}tX_n}]= \sum_{k=1}^{n}e^{\mathrm{i}t\frac{k}{n}}\cdot\frac{1} {n}= \frac{\exp\left(\mathrm{i}t\cdot\frac{n+1}{n}\right)-\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)}{\left(\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)-1\right)\cdot n} \rightarrow \frac{e^{\mathrm{i}t}-1}{\mathrm{i}t} \text{ as }n\rightarrow \infty$$