First In-Class Exam Solutions Math 410, Professor David Levermore Monday, 1 October 2018

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1 First In-Class Exam Solutions Math 40, Professor David Levermore Monday, October 208. [0] Let {b k } k N be a sequence in R and let A be a subset of R. Write the negations of the following assertions. (a) [5] For every M > 0 we have b j > M eventually as j. (b) [5] Every sequence in A has a subsequence that converges to a limit in A. Solution (a). There exists an M > 0 such that b j M frequently as j. Solution (b). There exists a sequence in A such that every subsequence of it either diverges or converges to a limit outside A. Remark. The answer There exists a sequence in A such that no subsequence of it converges to a limit in A. does not fully carry the negation through. Remark. Assertion (a) is equivalent to the sequence {b k } diverges to. Assertion (b) is the definition that the set A is sequentially compact. 2. [5] Give a counterexample to each of the following false assertions. (a) [5] If a sequence {a k } k N in R is bounded then it converges. b k+ (b) [5] If lim inf then b k diverges. b k k= (c) [5] A countable union of closed subsets of R is closed. Solution (a). A simple counterexample is the sequence {a k } k N with a k = ( ) k. It is bounded, but does not converge. Solution (b). A simple counterexample is the sequence {b k } k N with b k = k 2. This sequence satisfies ( ) b k+ k 2 2 lim inf = lim inf b k () = lim 2 + =, k but the series b k = k= k= k 2 converges, because it is the p-series with p = 2. Remark. This problem is asking for a convergent series about which the Ratio Test is inconclusive. There are many such examples! Any p-series with p > is one. Solution (c). A simple counterexample is the countable collection of closed intervals given by [0, 2 n ] for every n N. Each [0, 2 n ] is closed but their union [0, 2 n ] = [0, ) is not closed. n N

2 2 3. [0] Consider the real sequence { } k N given by = ( ) k 2k 3 for every k N = {0,, 2,... }. (a) [3] Write down the first three terms of the subsequence {c 2k } k N. (b) [3] Write down the first three terms of the subsequence {c 2k+ } k N. (c) [4] Write down lim inf and. (No proof is needed here.) Solution (a). When k = 0,, 2 we have 2k = 0, 2, 4, whereby the first three terms of the subsequence {c 2k } k N are c 0 = ( ) c 2 = ( ) c 4 = ( ) = 3, = 3, = 5 5 =. Solution (b). When k = 0,, 2 we have 2 =, 3, 5, whereby the first three terms of the subsequence {c 2k+ } k N are c = ( ) c 3 = ( ) c 5 = ( ) = 2 = 2, = 3 4, = 7 6. Solution (c). Because c 2k+ < 0 for k while c 2k > 0 for k, and because ( ) lim c 2k+ 2(2) 3 4k 2k+ = lim ( ) = lim (2) + 2k + 2 = 2, while we see that ( ) lim c 2k 2(2k) 3 2k = lim ( ) (2k) + lim inf = lim c 2k+ = 2, = lim 4k 3 2 = 2, = lim c 2k = 2.

3 4. [5] Let a 0 > 0 and define the sequence {a k } k N by a k+ = a k + 2 for every k N. (a) [0] Prove that {a k } k N converges. (b) [5] Evaluate lim a k. Solution (a). Notice that {a k } k N is a positive sequence. We will show that {a k } k N is also a contracting sequence, whereby it will be convergent. Notice that the recursion relation implies that for every k we have a k+ a k = a k + 2 a k + 2 ( ak = + 2 ) a k a k + 2 a k + 2 ak a k + 2 = a k a k ak a k + 2. Therefore, because {a k } k N is a positive sequence, we have for every k a k+ a k = a k a k ak a k + 2 < 2 2 a k a k. Because /(2 2) <, this implies that {a k } k N is a contracting sequence, whereby it is convergent. Solution (b). Let a R be the limit of the convergent sequence {a k } k N. By the recursion relation we have a 2 k+ = a k + 2. By letting k in this relation we see by the properties of limits that a 2 = a + 2, whereby either a = 2 or a =. Because {a k } k N is a positive sequence, we have lim a k = [0] Let A and B be any subsets of R. Prove that (A B) c A c B c. (Here S c denotes the closure of any S R.) Remark. We must show that every element of (A B) c is an element of A c B c. Solution. Let x (A B) c (be arbitrary). By the definition of closure there exists a sequence {x n } n N contained within A B such that x n x as n. Because {x n } n N is contained within A and x n x as n, we see that x A c by the definition of closure. Because {x n } n N is contained within B and x n x as n, we see that x B c by the definition of closure. Because x A c and x B c, we know that x A c B c. Because x (A B) c was arbitrary, we conclude that (A B) c A c B c. 3

4 4 6. [5] Let { } k N be a positive sequence in R. (a) [0] Prove that ck. (b) [5] Give an example for which the above inequality is strict. Solution (a). There is nothing to prove when =, so suppose that ρ = <. Because { } k N is a positive sequence, so is { / } k N, whereby ρ 0. Let r > ρ. By Proposition 2.7 we have < r eventually, say < r for every k m. Because > 0 for every k N and r > ρ 0, it follows by induction that Therefore which implies that c m r k m for every k m. ck r c m r m for every k m, ck r Because r > ρ was arbitrary, we have k cm r m r lim k c m r m = r. ck ρ = Therefore we have proved the desired inequality.. Solution (b). One example is given by = ( 3 ( ) k) k = { 4 k for k even, 2 k for k odd. It is clear that while Because 2 <, the inequality is strict. ck = lim 2k+ c 2k+ = 2, = lim c 2k+ c 2k = lim 4 2k 2 2k+ =.

5 7. [0] Let {b k } k N R be a sequence and {b nk } k N be a subsequence of it. Show that b k converges absolutely = b nk converges absolutely. Solution. By the definition of absolute convergence of a series b k converges absolutely b k converges, b nk converges absolutely b nk converges. By the definition of a convergent series, each of the series on the right-hand side above is convergent if and only if its associated sequence of partial sums is convergent. These sequences of partial sums are given by {q n } and {p m } respectively where q n and p m are defined for every n, m N by n m q n = b k, p m = b nk. It is clear that these sequences are nondecreasing. The Monotonic Sequence Theorem then implies that these sequences converge if and only if they are bounded above. Therefore b k converges {q n } is bounded above, b nk converges {p m } is bounded above. The crucial observation is that p m and q n satisfy the inequality m n m p m = b nk b k = q nm for every m N. This inequality shows that if {q n } is bounded above then {p m } is bounded above. Therefore b k converges absolutely {q n } is bounded above = {p m } is bounded above b nk converges absolutely. Remark. This proof involves three notions of convergence: () absolute convergence of a series, (2) convergence of a series, and (3) convergence of a sequence. Whenever converges appears in your solution it should be clear which notion is being used. 5

6 6 8. [5] Determine the set of all x R for which ( ) k 3 k x k converges. Give your reasoning. (The set is an interval. Be sure to check its endpoints!) Solution. Let a k denote the k th term in the sum, namely let We have a k+ a k = a k = ( ) k 3 k x k. 3 k+ x k+ = 3 x lim k k x k k + 2 = 3 x. The Ratio Test shows that the series converges absolutely for 3 x < and diverges for 3 x >. This test says nothing when 3 x =. When 3x = the series becomes ( ) k. Because the terms / are positive and decreasing with = 0, lim the Alternating Series Test can be applied to show that the series converges. When 3x = the series becomes. Because this is the p-series with p =, it diverges. Another argument is that because 2 the harmonic series diverges. and because for every k N, the Direct Comparison Test shows that the series diverges. Alternatively, because the terms / are positive and decreasing, the Integral Test or the Cauchy 2 k Test can be applied to show that the series diverges. Therefore the set of all x R for which the series converges is the interval (, 3 3]. Remark. It is not enough to argue that the series converges in the interval ( 3, 3 ]. You also have to argue that it diverges outside the interval.