(a) Prove that the sequence defined byyn= sup{ak:k≥n}converges.Proof.For anyn∈ {1,2,3, . . .}, define the setAn:={an, an+1, an+2, . . .}.2

In other words,yn= supAn. As defined,An+1⊂An, soyn+1≤yn.Since this is true for anyn, we see that the sequence (yn) is decreasing.Also, since (an) is bounded, there existsM>0 such that|an| ≤Mfor alln∈{1,2,3, . . .}. In particular, for anyn,-M≤an≤yn≤y1,so we see that the sequence (yn) is bounded.Therefore, since (yn) is a bounded, decreasing sequence, the Monotone ConvergenceTheorem implies that it converges.(b) Thelimit superiorof (an), or lim supan, is defined bylim supan= limyn,whereynis the sequence from part (a) of this exercise. Provide a reasonable definitionfor lim infanand briefly explain why it always exists for any bounded sequence.Answer.Here’s the definition of lim infan:For a bounded sequence (an), define the sequence (zn) byzn= inf{ak:k≥n}.Then thelimit inferiorof (an) is defined to belim infan:= limzn.The sequence (zn) is increasing and bounded (it’s bounded below byz1and above bythe upper bound for (an)), so the Monotone Convergence Theorem implies it converges,so the above limit definitely exists.(c) Prove that lim infan≤lim supanfor every bounded sequence, and give an example ofa sequence for which the inequality is strict.Proof.Note that, with (yn) and (zn) defined as above,zn≤an≤ynfor alln∈ {1,2,3, . . .}. In particular,zn≤yn. Therefore, by Theorem 2.3.4(ii),lim infan= limzn≤limyn= lim supan.Consider the sequence (an) given byan= 2 + (-1)n(1 + 1/n).Then it’s straightforward to show that lim infan= 1 and lim supan= 3, so the inequalityis strict for this sequence.3

(d) Show that lim infan= lim supanif and only if limanexists. In this case, all three sharethe same value.Proof.To show the forward implication, assume lim infan= lim supan. In other words,with (yn) and (zn) defined as above, limzn= limyn. Now, we know thatzn≤an≤ynfor alln∈ {1,2,3, . . .}, so the Squeeze Theorem implies that limanexists and is equalto the common limit of theyn’s andzn’s.Turning to the reverse implication, assume liman=Lfor someL∈R. Let>0. Since(an)→L, there existsN∈Nsuch that, whenevern≥N,|an-L|</2or, equivalently,-/2< an-L </2.

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- Math, lim, Mathematical analysis, Limit of a sequence, Limit superior and limit inferior