Final answer:
The sequence given by s1 = 0 and sn+1 = p × sn for n ≥ 1 is a geometric progression that converges to 0 if |p| < 1, since each term in the sequence decreases exponentially due to the common ratio p.
Step-by-step explanation:
The student is asked to prove the convergence of a given recursive sequence and to find its limit. The sequence is defined as s1 = 0 and for n ≥ 1, sn+1 = p × sn, where p is a constant factor. First, we note that such a recursive sequence forms a geometric progression with the first term being s1 = 0 and the common ratio being p.
To determine if this sequence converges, we must consider the value of p. A geometric sequence converges when its common ratio p has an absolute value less than 1 (i.e., |p| < 1). Assuming this is the case for this sequence, we can show that the limit as n approaches infinity for sn is 0 due to the factor p exponentially decreasing the value of each subsequent term in the sequence.
The limit of the sequence is calculated as:
ℒ sn = s1 × (1 - pn) / (1 - p) = 0 × (1 - 0) / (1 - p) = 0