Correct option is D
n=1∑∞(−1)n1+∣an∣an(i) Let an=n. Then:n=1∑∞(−1)n1+nn does not converge.So, Option A is incorrect.Now, let ank be a subsequence.If an is convergent, then:n→∞liman=n→∞limank.This implies:n→∞lim1+∣ank∣ank exists.If an diverges to ±∞ or oscillates, we can still find a number such that:n→∞limank=lorn→∞limank=±∞.In both cases:n→∞lim1+∣ank∣ank exists.Thus, we can say that for every sequence (an), there exists a number b such that:k=1∑∞b−1+∣ank∣ank converges.So, Option D is correct.