Proof: lim sup (Sn + tn) ≤ lim sup Sn + lim sup tn for Bounded Sequences

Since (Sn) and (tn) are bounded sequences, we know that there exist M1 and M2 such that |Sn| ≤ M1 and |tn| ≤ M2 for all n.

Let L1 = lim sup Sn and L2 = lim sup tn. Then, by definition of lim sup, we have:

• For any ε > 0, there exists N1 such that n > N1 implies Sn < L1 + ε.
• For any ε > 0, there exists N2 such that n > N2 implies tn < L2 + ε.

Now consider the sequence (Sn + tn). Since |Sn| ≤ M1 and |tn| ≤ M2, we have |Sn + tn| ≤ |Sn| + |tn| ≤ M1 + M2 for all n. This means that (Sn + tn) is also a bounded sequence.

Let L = lim sup (Sn + tn). Then, by definition of lim sup, we have:

• For any ε > 0, there exists N such that n > N implies (Sn + tn) < L + ε.

Now choose any ε > 0. Using the inequalities above, we can find N1 and N2 such that n > N1 implies Sn < L1 + ε/2 and n > N2 implies tn < L2 + ε/2. Then, for any n > max(N1, N2), we have:

(Sn + tn) < (L1 + ε/2) + (L2 + ε/2) = L1 + L2 + ε.

This shows that lim sup (Sn + tn) ≤ L1 + L2 + ε for any ε > 0, which implies lim sup (Sn + tn) ≤ L1 + L2. Therefore, we have:

lim sup (Sn + tn) ≤ lim sup Sn + lim sup tn.

As desired.