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Claim: Let {s n } be a bounded sequence. Let{t n } be a sequence such that t n 0

ID: 2937336 • Letter: C

Question

Claim: Let {sn} be a bounded sequence. Let{tn} be a sequence such that tn 0.Then {sntn} converges to zero also. Proof: Start by looking at |sntn- 0|< for an arbitrary . Simplifying to this|sn||tn|< {sn} bounded thus there exists an M such that M> |sn| So, |sn||tn| < M |tn|< Now want |tn| < (/M) We know there existsN such that for all n > N |tn-0|< (/M) forarbitrary . b/c sn is bounded . Case 2: if tn doenst converge to 0, what happens?This is where i need help! Suppose tn does converge to 0? Claim: Let {sn} be a bounded sequence. Let{tn} be a sequence such that tn 0.Then {sntn} converges to zero also. Proof: Start by looking at |sntn- 0|< for an arbitrary . Simplifying to this|sn||tn|< {sn} bounded thus there exists an M such that M> |sn| So, |sn||tn| < M |tn|< Now want |tn| < (/M) We know there existsN such that for all n > N |tn-0|< (/M) forarbitrary . b/c sn is bounded . Case 2: if tn doenst converge to 0, what happens?This is where i need help! Suppose tn does converge to 0?

Explanation / Answer

Suppose tn doesn't converge then the sequencesntn need not be a convergentsequence. For example sn = (-1)n thenclearly it is a bounded sequence.           Taketn = 1+1/n. Then tn is a sequence whichconverges to 1. Then  sntn =(-1)n(1+1/n) .         It is easy to checkthat  sntn is not a convergentsequence.