Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 26K / b / i / Solution

For $n>m$, independence and $\mathbb{E}{X}_k=0$ giveThus $(S_n)$ is a Cauchy sequence in the complete L2 space and converges in $L^2$ to some random variable $S$. The Cauchy-Schwarz inequality gives $\Vert Z{\Vert }_1\le \Vert Z{\Vert }_2$ on a probability space, so the convergence also holds in $L^1$. Finally, convergence in $L^2$ implies convergence in probability, which implies convergence in distribution.