Question

Prove or give a counterexample. (a) If sₙ → s and sₙ > 0 for all n, then s > 0. (b) If {sₙ} and {tₙ} are divergent sequences, then {sₙ + tₙ} is divergent. (c) If {sₙ} and {tₙ} are divergent sequences, then {sₙtₙ} is divergent. (d) If {sₙ} and {sₙ + tₙ} are convergent sequences, then {tₙ} is convergent. (e) If {sₙ} and {sₙtₙ} are convergent sequences, then {tₙ} is convergent. (f) If {sₙ} is not bounded above, then {sₙ} diverges to +infinity.

Answer 1

(a) If sₙ → s and sₙ > 0 for all n, then s > 0

Step-by-step explanation:

(a) If sₙ → s and sₙ > 0 for all n, then s > 0:

To prove this, we can use the fact that the limit of a sequence is unique. Since sₙ → s, we know that for any positive ε, there exists an N such that for all n > N, |sₙ - s| < ε. Let ε = s/2. Then, for n > N, we have |sₙ - s| < s/2, which implies -s/2 < sₙ - s < s/2. Adding s to all sides gives s > sₙ - s/2, which simplifies to s/2 < sₙ. Since s/2 is positive, we conclude that s > 0.