Correct answer: (A) X_n→X in distribution.; (B) X_n→X in probability.; (D) Some subsequence X_(n_k)→X almost surely.
Explanation
For fixed ε>0 and sufficiently large n,P(|X_n|>ε)=1/(n+1)→0. This proves B and therefore A. The means n/(n+1)→1≠0,so C fails. Choose n_k=2^k;the exceptional probabilities sum to a finite value,so Borel–Cantelli proves almost-sure subsequence convergence. **A,B,D**.
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