Math 314 Test Two Review Problems (Nowwith Solutions!)Dr. HolmesThese are problems to study for the test. Some of these may be testquestions or close variations of test questions. I am much more likely toanswer these in detail during office hours than during class on Wednesday. Iwill hold TTh 3-5 office hours this week in addition to my usual hours.I may post answers to some of these (but probably not all of them) onThursday some time.Keep watching this document: I may add more problems.Wednesday I added a couple of problems!1. New part added to this question(a) Show thatlim sup(sn+ tn) = lim sup sn+ lim sup tnis NOT a theorem by presenting specific sequences (sn) and (tn)for which this is not true. Periodic sequences make good examples.Give an example where (sn+ tn) is not a constant sequence.Solution: Consider 1,2,3,1,2,3,. . . and 1,3,2,1,3,2. . . The lim supof each of these sequences is 3 but the lim sup of the sum sequence2,5,5,2,5,5. . . is 5 < 3 + 3.(b) Prove thatlim sup(sn+ tn) ≤ lim sup sn+ lim sup tn.Of course this is familiar.Solution: I proved this in class.1Define a1as 2 and an+1as an(1 −1(n+1)3) for each natural number n.Explain why this sequence must converge, justifying all your claims asfar as possible.Since 0 < 1 −1(n+1)3< 1, we have an< an(1 −1(n+1)3) = an+1for eachn, so the sequence is nonincreasing (in fact, it is strictly decreasing).Further, notice that a1is positive, and if we suppose that anis positive,and notice that 1−1(n+1)3is positive, it follows that an(1−1(n+1)3) = an+1is positive, and so by mathematical induction anis positive for everyn, so the sequence is bounded below (by 0).A nonincreasing sequence which is bounded below must converge.2.3. Define a1= 1; an+1= 1 +12an.Prove that an< 2 for all n by induction on n.Solution: Clearly a1< 2.Suppose ak< 2. Our goal is to prove ak+1< 2.ak+1= 1 +12akwhich is less than 1 +12(2) by the ind hyp and obviousproperties of order, and 1 +12(2) = 2, which completes the proof byinduction.Prove that this is an increasing sequence.Solution: What we need to prove is that an< an+1for each n.Basis step: a1< a2: 1 <32is obvious.Induction step: Suppose that ak< ak+1. Our goal is to prove ak+1<ak+2. ak+1= 1 +12ak, which is less than 1 +12ak+1by ind hyp andobvious properties of order, and this is equal to ak+2, completing theproof by induction.Explain why the last two results show that the sequence must converge;then use limit theorems and the fact that lim an+1= lim anto evaluatethe limit.Solution:The sequence is shown to be nondecreasing and bounded above: so bya well known theorem it must converge.2So we know that lim an= L for some real number L. So lim an=lim an+1= lim(1+12an) = lim 1+12lim an(by limit theorems) = 1+12L.From L = 1 +12L we deduce12L = 1 and L = 2. So lim an= 2.4. State the definition of a Cauchy sequence. Explain in detail why sn=(−1)nis not a Cauchy sequence (from the definition, with due attentionto logic, not from any theorems you may happen to know about Cauchysequences).Solution: A sequence (sn) is a Cauchy sequence iff for each  > 0there is N such that for all natural numbers m, n such that m > Nand n > N, we have |sm− sn| < .To see that (−1)nis not a Cauchy sequence, set  = 1. Choose any N.There is an even natural number 2n > N and of course 2n + 1 > Ntoo. Now |(−1)2n− (−1)2n+1| = |1 − (−1)| = 2 > 1, so we cannot findany N such that for all m, n > N we have |(−1)m− (−1)n| < 1.5. Prove that a nondecreasing sequence (sn) such that {sn| n ∈ N} isbounded above must converge to a real number. (This is in the book:familiarize yourself with this proof).6. In this question, we develop the definition of lim sup snand the proofthat lim sup snwill exist and be a real number if {sn| n ∈ N} isbounded above.Define the sequence (un) of which lim sup snis limit and then state thedefinition of lim sup sn(without using the notation unin the definition).Solution: un= sup{sN| N > n}Assume that {sn| n ∈ N} is bounded (above and below: this is acorrection of the way the problem was stated initially). Explain whatproperties (un) has (and why it has them) which ensure that the limitof this sequence must exist.Solution: The sequence unis nonincreasing. If m > n, then umisthe supremum of {sN| sN> m} which is a proper subset of the set{sN| sN> n} of which unis supremum. So any upper bound ofthe larger set is also an upper bound of the smaller set, so unis anupper bound of {sN| sN> m}, so un≥ um. Because we have addedthe assumption that snis bounded below, we can conclude that un3is bounded below (by any lower bound for all the sn’s) and so mustconverge.7. Let (sn) be a sequence with lim sup sn= L, a real number. Prove thatfor each M > L there is an N such that for all n > N , sn< M.Solution: L = lim sup sn= limN →∞sup{sn| n > N }.By the definition of limit, for some N2, for all N > N2, sup{sn| n >N} < L + , where  = M − L, so sup{sn| n > N} < M for allN > N2. But from this and definition of the sup it follows immediatelythat sn< M for all n > N2+ 1: sup{sn| n > N2+ 1} < M, so everyelement of the set {sn| n > N2+ 1} is less than M .It’s N2+ 1 that witnesses the truth of the statement of course; weare fighting here with the fact that N was already used as a dummyvariable.Explain why it might not be true that there is an N such that for alln > N, sn< L (the result above with M = L instead of M > L).Draw a picture of a sequence for which this is not true.Solution: Think about the sequence 1 +1n. Any sequence that con-verges to a limit from above is an example.8. Suppose that we can show thatlim inf sn≤ lim inf tn≤ lim sup tn≤ lim sup sn.Suppose that snconverges. Explain in detail, stating and using theo-rems relating limit, lim sup, and lim inf, why tnmust converge.If sn, then lim inf sn= lim sup snby a theorem. By the inequalitiesabove, lim inf tn= lim sup tnis forced to be true. It then followsthat lim tnexists. The theorem says that lim snexists iff lim inf sn=lim sup snand moreover lim sn= lim inf sn= lim sup snwill be true inthis case.9. Suppose that (rn) is a sequence in which each rational number in theinteral [0,1] appears once. Prove that (rn) has a subsequence rσ(n)which converges to12. (This is almost exactly like 11.7).4Solution: Choose σ(1) so that rσ(1)=12. (It really doesn’t matterwhat I pick: I can do this because r enumerates all the rationals in theinterval).The condition I want to enforce on σ is that it is an increasing