Midterm Review Math 311, Spring 2016

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1 Midterm Review Math 3, Spring 206 Material Review Preliminaries and Chapter Chapter 2. Set theory (DeMorgan s laws, infinite collections of sets, nested sets, cardinality) 2. Functions (image, preimage, injection, surjection, bijection) 3. The absolute value function on R and its properties (e.g. triangle inequality) 4. Well-ordering property of N and Principle of Mathematical Induction 5. Properties (axioms) of fields and ordered fields; in particular, for R and Q. 6. Axiom of Completeness; infimum, supremum of a set of real numbers 7. Nested Interval Property; Archimedean Property; Irrational numbers; Density of Q. Sequences; a sequence is a function f : N R (f (n) = x n ); a countable set is the range of a sequence of distinct terms 2. x n x if for every ε > 0 there is a K(ε) > 0 such that x n x < ε for all n > K(ε) 3. Convergent sequences are bounded; Algebraic Limit Theorem, Order Limit Theorem 4. Definition of monotone sequences; Monotone Convergence Theorem 5. Definition of subsequences; x n x x nk x for every subsequence of x n 6. Divergence criteria (x n unbounded; any two subsequences of x n converge to distinct limits) 7. Every sequence of real numbers has a monotone subsequence; Bolzano-Weierstass Theorem (a bounded sequence of real numbers has a convergent subsequence) 8. Cauchy sequences; prove sequence is Cauchy using definition; Cauchy sequences are bounded; a sequence of real numbers converges if and only if it is Cauchy 9. Infinite series; convergence; divergence; absolute convergence; Cauchy criterion for convergence of series; if x k converges then limx k = 0; a series of nonnegative terms converges if and only if the partial sums {s n } form a bounded sequence; Comparison test; Geometric series; Cauchy Condensation Theorem; /n p converges if and only if p >

2 True/False (T/F). a) If a,b R and a < b, then there is an irrational number r such that a < r < b. b) If f : A B is injective and g : B C is surjective, then g f : A C is injective. c) If x n x and {x nk } is a subsequence, then x nk x. d) If x n x, then x n x. e) If A R is bounded, then there exists a closed, bounded interval I such that A I. Sample Problems P. For the set A = { n m n,m N}, determine infa,supa. P2. Determine a bijection between N and a proper subset of N. P3. Determine a bijection between N and the set of all odd integers greater than 7. P4. Prove the following statements. a) If A, B R are bounded subsets of R, then sup(a B) = sup{sup A, sup B}. b) Let {x n } n= be a sequence such that x n x. Suppose x > a. Then there exists a number N such that x n > a for all n > N. c) For any x R, there exists a sequence {q n } of rational numbers, which is strictly increasing and q n x. d) If {x n },{y n } are Cauchy sequences, then { x n y n } is a Cauchy sequence. e) If {x n } is unbounded, there exists a subsequence {x nk } such that lim x nk = 0. f) Show that a bounded, monotone increasing sequence is Cauchy, using the definition. g) Let a > 0 and x > 0. Inductively define x n+ = a + x n for n N. Prove {x n } converges and determine the limit. h) (Bernoulli s Inequality) If x >, then ( + x) n + nx, for all n N. i) Prove that n=3 n(logn)(loglogn) p

3 converges if p >. j) Prove diverges. cosn n= Solutions Solution 4c) Let x R (which is fixed). We want to construct an increasing sequence {q n } of rational numbers such that q n x. For any n N, x n < x + n. By the density of Q in R (for any a,b R with a < b there exists q Q such that a < q < b), there exists q n Q such that x n < q n < x + n. This inequality is equivalent to q n x < n. By the Archimedean property, for any ε > 0 there exists n N such that 0 < n < ε, hence q n x < n < ε. This shows that there is some sequence {q n } of rational numbers which converges to x. This sequence may or may not be monotonically increasing. However, every sequence of real numbers has a monotone subsequence. Now, the question is, can we necessarily determine a subsequence that is monotonically increasing? If yes, then how can we do so? If not, then how can this argument be changed to yield a sequence/subsequence that is monotonically increasing? Solution 4d) Assume {x n },{y n } are both Cauchy sequences. For any ε > 0, there exist K (ε),k 2 (ε) such that x n x m < ε, for n,m > K (ε) and y n y m < ε, for n,m > K 2 (ε). We want to show that the sequence {z n } with terms z n = x n y n is also Cauchy. By the backwards triangle inequality, we have z n z m = xn y n x m y m (x n y n ) (x m y m ) = (x n x m ) + (y m y n ) x n x m + y m y n. (0.)

4 We need to choose a suitable N(ε), such that z n z m < ε for n,m > N(ε). If n,m > K (ε/2) and n,m > K 2 (ε/2), then each term in (0.) will be bounded by ε/2. Thus, the conclusion will follow if we take N(ε) = max{k ( ε 2 ),K 2( ε 2 )}. Solution 4e) Suppose {x n } is not bounded above. This means, for every M > 0 there exists a term in the sequence which is greater than (or equal to) M. So, for every M > 0, there exists m N such that x nm > M. Then, by definition, the subsequence {x nm } diverges to. Note that, since M is positive, every term of the subsequence is positive, which may not be true for the original sequence. Since x nm > 0 for all n m, x nm diverges to if and only if x n m 0. Solution 4f) Suppose {x n } is a bounded, monotonically increasing sequence. Since {x,x 2,...} is a bounded set, the supremum exists. Let s = sup{x,x 2,...}. Then, for every ε > 0 there exists k N such that s ε < x k. Since the sequence is increasing, for all n > k we have x k x n, so s ε < x k x n < s < s + ε. This inequality implies x n x k < ε for all n > k. To see this, it may help to separate s ε < x k x n < s < s + ε into two inequalities, namely s ε < x n < s s < x k < ε s. Solution 4g) This problem is a little bit too cumbersome for the in-class midterm, as many would likely be inclined to consider cases. However, we can argue that this sequence is monotonically increasing, given that both a and x are positive. By definition, x 2 = a + x, so x 2 x = a + x x = a + x x 2 a + x + x. (0.2) The denominator is positive, since both a and x are positive. The numerator can be shown to be strictly positive as well (the numerator is a quadratic; see what it would take for it to be negative). This shows that x 2 x 0 which implies x 2 x. For induction, assume x n x n. Then x n+ = a + x n = x 2 n+ = a + x n a + x n = x 2 n, hence x 2 n+ x2 n = x n+ x n. By induction, x n+ x n for all n N, so {x n } is monotonically increasing. Since x > 0, this sequence is bounded below by 0. By examining the quadratic x 2 + x + a in the numerator of (0.2), one can arrive at the potential upper bound (and potential limit) of 2 ( + + 4a).

5 Solution 4i) The function n(logn)(loglogn), appearing in the denominator of the series is increasing, hence n=3 n(logn)(loglogn) p, (0.3) n(logn)(loglogn) p is decreasing. Thus, we are able to apply the Cauchy Condensation Test. The series (0.3) converges if and only if the series converges. The series (0.4) simplifies to Now, log(log 2) < 0, so The series = 2 k 2 k (log2 k )(loglog2 k ) p (0.4) (k log2)(log(k log2)) p (k log2)(logk + loglog2) p. (0.5) (k log2)(logk + loglog2) p k(logk) p. k(logk) p should be familiar from class, so the conclusion should follow. I have left out some details that you should fill in. Clearly, this problem is too long and involved for the in-class midterm. Solution 4j) To show that cosn n= diverges, it is enough to show that limcosn 0. There are various ways to show this, depending on the level of rigor desired. As a sequence, one could argue that {cosn} has two subsequences that converge to distinct limits. In particular, since cosn, it

6 should not be terribly hard to argue that some subsequence converges to, while another subsequence converges to. This can be accomplished directly or by using the notions of limit supremum and limit infimum (Exercise 2.4.7; more on these topics later in class). One can also show that {cosn} diverges by contradiction. I will outline one possible argument by contradiction. Suppose cosn L as n, then cos(n + 2) converges to L (since any subsequence converges to the same limit). Applying various trigonometric identities, it can be show that cos(n + 2) cos n = 2 sin() sin(n + ). Letting n, this shows that sinn 0. Then, in the limit as n, numerous trigonometric identites will be contradicted, such as the Pythagorean theorem.