 Solution

Almost sure convergence implies convergence in probability, so part (b) identifies the limit as a constant $c$ almost surely. For each $ϵ>0$, the events $A_n=\{|X_n-c|>ϵ\}$ are independent. Almost sure convergence gives $\mathrm{Pr}(A_n\text{ infinitely often})=0$. If their probability sum diverged, the second Borel-Cantelli lemma would instead give probability one. Hence the sum is finite for every $ϵ$, proving almost sure convergence implies complete convergence for independent sequences.