How to prove the weak convergence of random variable $\sum_{i=1}^{n} \quad X_n$: $$P(X_n=0)=P(X_n=\frac{1}{2^n})=\frac{1}{2}$$ to the random variable with uniform distribution. $X_1,...,X_n$ are independent.
I tried from the characterisitc function and i don't know how to calculate: $$(\frac{1}{2}e^{it\frac{1}{2}}+\frac{1}{2}) \cdot ... \cdot (\frac{1}{2}e^{it\frac{1}{2^n}}+\frac{1}{2})$$
Thanks in advance.
$$\begin{aligned} \phi_n(t) &= \prod_{k=1}^n E(e^{itX_k})\\ &= \prod_{k=1}^n \frac 12 e^{it\frac{1}{2^k}} + \frac 12\\ &= \prod_{k=1}^n \frac 12 e^{it\frac{1}{2^{k+1}}}(e^{it\frac{1}{2^{k+1}}}+e^{-it\frac{1}{2^{k+1}}})\\ &= \prod_{k=1}^n \frac 12 e^{it\frac{1}{2^{k+1}}}2\cos\left(\frac{1}{2^{k+1}}\right)\\\ &=\exp(it \sum_{k=1}^n \frac{1}{2^{k+1}}) \prod_{k=1}^n \cos\left(\frac{1}{2^{k+1}}\right)\\ &= \exp\left(it\frac12 \left(1-\frac 1{2^n} \right)\right) \frac{\sin(t)}{2^{n+1}} \frac{1}{\sin\left(\frac t{2^{n+1}} \right)}\frac{1}{\cos\left(\frac t2\right)} \\ &\to e^{i\frac t2}\frac{\sin(t)}{t\cos\left(\frac t2\right)}\\ &= e^{i\frac t2} \frac{\sin\left(\frac t2\right)}{\frac t2}\\ &= \frac{e^{it}-1}{it} \end{aligned}$$
$\phi_n$ converges pointwise to the c.f. of $\mathcal U([0,1])$, hence convergence in distribution.