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