Prove lim sup |sn| = 0 if and only if lim sn = 0

We will prove the statement using the following two lemmas:

Lemma 1: If lim sup an = 0, then there exists a subsequence of (an) that converges to 0.

Proof: Let ε > 0 be arbitrary. By definition of lim sup, there exists an integer N such that for all n ≥ N, we have |an - lim sup an| < ε. In particular, this implies that for all n ≥ N, we have an < lim sup an + ε. Since lim sup an = 0, we can choose ε small enough so that lim sup an + ε < ε. Therefore, for all n ≥ N, we have an < ε, which means that the subsequence (an)N∞ converges to 0.

Lemma 2: If lim sn = 0, then lim sup |sn| = 0.

Proof: Let ε > 0 be arbitrary. Since lim sn = 0, there exists an integer N such that for all n ≥ N, we have |sn| < ε/2. Let M = max{|s1|, |s2|, ..., |sN-1|}. Then for all n ≥ N, we have

|sn| ≤ |s1| + |s2| + ... + |sN-1| + |sn| ≤ M + |sN| + |sN+1| + ...

Therefore, for all n ≥ N, we have |sn| ≤ (M + |sN| + |sN+1| + ...) + ε/2. Since the sum |sN| + |sN+1| + ... is finite, there exists an integer K such that for all n ≥ K, we have |sN| + |sN+1| + ... + |sn| < ε/2. Therefore, for all n ≥ max{N, K}, we have |sn| < (M + ε/2) + ε/2 = M + ε. Since M is a fixed constant, this implies that lim sup |sn| ≤ M + ε. Since ε was arbitrary, we have lim sup |sn| ≤ M, which means that lim sup |sn| = 0.

Now we can prove the original statement. Suppose lim sup |sn| = 0. By Lemma 1, there exists a subsequence (In) of (|sn|) that converges to 0. Since |sn| ≤ In for all n, this implies that lim In = 0. By definition of |sn|, we have sn = |s1| + ... + |sn| ≤ |I1| + ... + |In| for all n. Therefore, lim sup sn ≤ lim sup (|I1| + ... + |In|) = lim (|I1| + ... + |In|) = 0, where the last equality follows from the fact that (In) converges to 0. This proves that lim sn = 0.

Conversely, suppose lim sn = 0. By Lemma 2, we have lim sup |sn| = 0. Therefore, the statement is proved.