Let snand tnbe bounded sequences of nonnegative numbers Prove lim sup sntn lim sup sn lim sup tn

Since (sn) and (tn) are bounded sequences, we have:

0 ≤ sn ≤ M1 for some positive number M1, and 0 ≤ tn ≤ M2 for some positive number M2.

Therefore, for any ε > 0, we can choose n1 and n2 such that:

sn ≤ (lim sup sn) + ε/M2 for all n ≥ n1, and tn ≤ (lim sup tn) + ε/M1 for all n ≥ n2.

Choose N = max{n1, n2}. Then for all n ≥ N, we have:

sn ≤ (lim sup sn) + ε/M2 and tn ≤ (lim sup tn) + ε/M1.

Multiplying the two inequalities, we get:

sn tn ≤ ((lim sup sn) + ε/M2)((lim sup tn) + ε/M1)

Expanding the right side and simplifying, we get:

sn tn ≤ (lim sup sn)(lim sup tn) + ε(lim sup sn)/M1 + ε(lim sup tn)/M2 + ε^2/(M1M2)

Taking the lim sup of both sides as n → ∞, we get:

lim sup (sn tn) ≤ (lim sup sn)(lim sup tn) + ε(lim sup sn)/M1 + ε(lim sup tn)/M2 + ε^2/(M1M2)

Since ε > 0 was arbitrary, we can let ε → 0 and obtain:

lim sup (sn tn) ≤ (lim sup sn)(lim sup tn)

Therefore, we have:

lim sup (sn tn) < (lim sup sn)(lim sup tn)

since the inequality is strict (we used ≤ in the proof)