Let $(X_n)_n$ be a sequence of independent random variable. Let $W_n=\sum_{k=1}^nX_k.$ We suppose that $(W_n)_n$ converges in distribution.
Prove that $(W_n)_n$ also converges almost surely and in probability.
I know that if $(W_n)_n$ converges in probability then it converges almost surely (for the proof, one can use Etemadi's maximal inequality), so it is sufficient to have the probability convergence.
there is a hint, but I don't know if it is helpful, we can use $e^{itW_n}$ and the characteristic functions.
What ways do we have have to prove this result? And is it easier to prove, first, the almost sure convergence?
Hint: if $n_k$ and $m_k$ increase to $\infty$ then $\prod_{j=n_k}^{m+k} Ee^{it X_j} \to 1$ for all $t$ because $\prod_{j=1}^{\infty} Ee^{it X_j} $ is the characteristic function of the limit of $(W_n)$ in distribution. Hence $W_{n_k} -W_{m_k} $ tends to $0$ in distribution which implies convergence in probability. This implies that $\{W_n\}$ is Cauchy in probability and hence it converges in probability.