This exam contains 5 pages (including this cover page) and 4 questions. The total number of points is 100. Grade Table

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1 MAT25-2 Summer Session Practice Final August 24th, 207 Time Limit: Hour 40 Minutes Name: Instructor: Nathaniel Gallup This exam contains 5 pages (including this cover page) and 4 questions. The total number of points is 00. Grade Table Question Points Score Total: 00

2 Problem (25 points). (a) Let (a n ) be a bounded sequence. Define lim sup a n and lim inf a n. (b) Prove that lim sup a n b if and only if for all ɛ > 0 there exists N N such that for all n N, we have a n < b + ɛ. (c) Prove that there exists a subsequence of (a n ) that converges to lim sup a n. Solution. (a) Define x n = sup{a k k n} and y n = inf{a k k n}. Then lim sup a n = lim n x n and lim inf a n = lim n y n. (b) Let l = lim sup a n. ( ). Suppose that l b, and let ɛ > 0 be arbitrary. Then by definition of convergence of (x n ) l, there exists N N such that for all n N, we have x n l < ɛ, which implies that x n l < ɛ. Adding l to both sides, and noting that l b by hypothesis, yields x n < l + ɛ b + ɛ for all n N. But since x n = sup{a k k n}, it follows that a n < b + ɛ for all n N, as desired. ( ). Suppose that for all ɛ > 0 there exists N N such that for all n N, we have a n < b + ɛ, and suppose for contradiction, that l > b. Then let ɛ = (b + l)/2 > 0. Then by hypothesis there exists N N such that if n N, then we have a n < b + ɛ = b + (b + l)/2. In particular this means that x N = sup{a n n N} b + (b + l)/2 < l. However (x n ) is a decreasing sequence which converges to l, and therefore x n l for all n N. So we obtain a contradiction, and it follows that l b, as desired. (c) Let l = lim sup a n. Since x n = sup{a k k n} and (x n ) l, we can use axiom (RS2) for each k N, to inductively choose a nk with n k n k, and x k k < a n k x k. Since n k n k, it follows that (a nk ) is a subsequence of (a n ), and by the squeeze theorem, since (x k /k) l and (x k ) l, we have that (a nk ) l as well. 2

3 Problem 2 (25 points). (a) State what it means for a series n= x n to converge to x R. (b) State the Alternating Series Test, the comparison test, and the absolute convergence test. (c) Does the series ( ) k+ k converge absolutely, conditionally, or diverge? Prove your result. (d) Prove or find a counterexample: If n= a n converges, then (d) Prove or find a counterexample: If n= a n converges absolutely, then Solution 2. (a) A series n= x n converges to x R if its sequence of partial sums (s m ), defined by s m = m n= x n, converges. 3 (b) Theorem : Alternating Series Test (Abbott Theorem 2.7.7) Let (a n ) be a sequence satisfying the following statements. (a) a a 2 a (b) (a n ) 0. Then the series ( ) n+ a n = a a 2 + a 3 a 4 + a 5... converges. n= Theorem 2: Comparison Test (Abbott Theorem 2.7.4) Let (a k ) and (b k ) be sequences with 0 a k b k for all k N. (a) If b k converges, then a k converges. (b) If a k diverges, then b k diverges. Theorem 3: Absolute Convergence Test (Abbott Theorem 2.7.6) If a series converges absolutely, then it converges. ( ) (c) This series converges conditionally. Since k is a positive, decreasing sequence which converges to 0, the alternating series ( ) k+ k converges by the Alternating Series Test. However it does not converge absolutely, since ( ) k+ k = k = k /2, which is a p-series with p <, and therefore it diverges. (d) This is not true. For example, consider the series from the previous parts of this problem ( ) k+ k. This converges, but ( ) ( ) k+ 2 k = k is the harmonic series, which diverges. (e) This is true. If n= a n converges, then the sequence of terms ( a n ) n= converges to 0. This implies that there is some N N such that for all n N, we have a n <. Therefore, for all n N, we have a 2 n = a n 2 < a n. So by the comparison test, it follows that because n= a n converges, we have that

4 Problem 3 (25 points). Let (a n ) be a sequence of distinct points, i.e. a n a m for n m. (a) State the definition of a closed set F R and a limit point of a set A R. (b) State the limit point characterization of a closed subset of R. (c) If (a n ) converges to a R, show a is the unique limit point of the set {a n n N}. (d) If (a n ) a, show that {a n n N} {a} is closed. (e) Give an example of a sequence (b n ) with two distinct limit points x, y R. Solution 3. (a) Definition : Closed Set A set F R is closed if F c = R F is open. Definition 2: Limit Point (Abbott Definition 3.2.4) Let A be a subset of R. A point x R is a limit point of A if every ɛ-neighborhood V ɛ (x) of x intersects A at some point other than x, i.e. for all ɛ > 0, there exists some y x with y V ɛ (x) A. 4 (b) Theorem 4: Limit Point Characterization of Closed Sets (Abbott Theorem 3.2.3) A set F R is closed if and only if it contains all of its limit points. (c) Since the terms of (a n ) are distinct, at most one of them is equal to a. Therefore, there exists a subsequence (a nk ) of (a n ), which we may obtain by removing either one or none of the terms of (a n ), such that none of the terms of this subsequence are equal to a. Then since subsequences converge to the same limit as their parent sequences, (a nk ) a as well. Therefore a is a limit of a sequence in {a n n N}, so a is a limit point of this set. (d) Since the only limit point of {a n n N} is a, it follows that {a n n N} = {a n n N} {a}, and this set is closed since the closure of any set is closed. (e) Define (b n ) = (, 2, /2, + /2, /3, + /3,...). Then the sequence (, /2, /3,...) is contained in {b n n N}, and this sequence is never equal to 0, yet converges to 0, so 0 is a limit point of {b n n N}. Additionally, the sequence ( +, + /2, + /3,...) is contained in {b n n N}, and this sequence is never equal to, yet converges to, so is a limit point of {b n n N} as well.

5 Problem 4 (25 points). (a) State the definition of an open set O R. (b) Can an open set contain one of its limit points? Prove or find a counter-example. (c) Let A R be a nonempty, bounded subset. Show that if sup A / A, then sup A is a limit point of A. (c) Show that if U is an open, nonempty, bounded subset of R, then sup U / U. (d) Show that if F is a closed, nonempty, bounded subset of R, then sup F F. Solution 4. (a) Definition 3: Open Set (Abbott Definition 3.2.) A set O R is open if for every a O there exists an ɛ-neighborhood (??) V ɛ (a) of a such that V ɛ (a) O. (b) Yes, an open set can contain one of its limit points. In class we showed that open intervals are open sets, hence (0, ) is an open set. Furthermore, we showed that the set of limit points of (0, ) is [0, ], hence /2 (0, ) is a limit point of (0, ), as desired. (c) Since A is nonempty and bounded, it has a supremum. Let s = sup A. Then given any ɛ > 0, by (RS2), there exists some a A with s ɛ < a. This implies that a V ɛ (s). Since a A and s / A, it follows that s a. Therefore we have shown that for every ɛ > 0, there exists some a V ɛ (s) A with a s, showing that s is a limit point of A. (c) Let U be an open, nonempty, bounded subset of R. Then s = sup U exists. If s U, then because U is open, there exists some ɛ > 0 such that V ɛ (s) U. However s < s + ɛ/2, yet s + ɛ/2 is contained in V ɛ (s), and therefore in U, contradicting that s is an upper bound for U. Therefore s / U, as desired. (d) Let F be a closed, nonempty, bounded subset of R. The s = sup F exists. Suppose, for contradiction, that s / F. Then by part (c), it follows that s is a limit point of F. However in class we showed that closed sets contain all of their limit points. Therefore s F, which is a contradiction. It follows that in fact s F. 5