MAT25 LECTURE 10 NOTES. = a b. > 0, there exists N N such that if n N, then a n a < ɛ

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1 MAT5 LECTURE 0 NOTES NATHANIEL GALLUP. Algebraic Limit Theorem Theorem : Algebraic Limit Theorem (Abbott Theorem.3.3) Let (a n ) and ( ) be sequences of real numbers such that lim n a n = a and lim n = b. Then the following statements hold. (a) lim n (ca n ) = ca. (b) lim n (a n + ) = a + b. (c) lim n (a n ) = ab. (d) If 0 for all n and b 0, then lim n a n bn = a b. Proof. (a) We consider two cases. (c 0). (Scratch Work). We are trying to show that (ca n ) ca. Hence we are interested in the inequality (b) ca n ca < c(a n a) < a n a < a n a <. (Proof ). Let > 0 be arbitrary. Since (a n ) converges to a, given > 0, there exists N N such that if n N, then a n a <. Then we compute a n a < = a n a < = c(a n a) < = ca n ca < Therefore for any > 0, we have found N N such that if n N, then ca n ca <. Hence (ca n ) converges to (ca). (c = 0). If c = 0, then the sequence (ca n ) is simply the sequence (0, 0, 0,...) which consists only of 0 s. By??, this sequence converges to 0 = ca, as desired. (Scratch Work). We are trying to show that (a n + ) a+b. Hence we are interested in the quantity (a n + ) (a + b) = (a n a) + ( b) () a n a + b Here () follows from the triangle inequality (??). But since (a n ) a and ( ) b, we can make a n a and b as small as we want. (Proof ). Let > 0 be arbitrary. Since (a n ) a, there exists some N N such that if n N, we have that a n a <. Similarly, since () b, there exists some N N such that if n N, we have that b <. Let N = max{n, N }. Then we have that N N and N N, so if n N, then n N and n N, so we have both a n a < and b <. Compute

2 MAT5 LECTURE 0 NOTES (c) (a n + ) (a + b) = (a n a) + ( b) a n a + b < + Therefore for any > 0, we have found N N such that if n N, then (a n + ) (a + b) <. Hence (a n + ) converges to (a + b). (Scratch Work). We are trying to show that (a n ) ab. Hence we are interested in the quantity a n ab = a n a + a ab () a n a + a ab = a n a + a b. Here () follows from the triangle inequality (??). Since (a n ) a and ( ) b, we can make a n a and b small. The number a is fixed, and since ( ) is a convergent sequence,?? implies that ( ) is bounded. Hence we have some bound M R with M for all n N. (Proof ). Let > 0 be arbitrary. Since ( ) is a convergent sequence,?? implies that ( ) is bounded. Hence we have some bound M R, which we can choose to satisfy M > 0, with M for all n N. Now we consider two cases. (a = 0). Since (a n ) a, there exists some N N such that if n N, we have a n a < M. Then compute a n ab = a n a + a ab a n a + a ab = a n a + a b < M M + 0 (a 0). Since (a n ) a, there exists some N N such that if n N, we have a n a < M. Since ( ) b and a 0, there exists some N N such that if n N, we have b < a. Let N = max{n, N }. Then we have that N N and N N, so if n N, then n N and n N, so we have both a n a < M and b < a. Compute a n ab = a n a + a ab a n a + a ab = a n a + a b < M M + a a Therefore in either case for any > 0, we have found N N such that if n N, then (a n ) (ab) <. Hence (a n ) converges to (ab). ( ) (d) First we show that if 0 for all n N, then b ( ). (Scratch Work). We are trying to show that b. Hence we are interested in the quantity b = b b. We need to make the quantity b bn b small. Since ( ) b, and b is a fixed number, we can make b b small, but to make small, we need a lower bound on the terms. This is make possible by the fact that ( ) b and b is nonzero, hence the terms of are eventually bounded away from 0.

3 MAT5 LECTURE 0 NOTES 3 (Proof ). Let > 0 be arbitrary. Since ( ) b, there exists N N such that if n N, then b < b. Using the reverse triangle inequality (??), we obtain that b b < b = b < b = 0 < b < () = < b. Here () follows from?? part (e). Furthermore, because ( ) b, there exists N N such that if n N, then b < b. Let N = max{n, N }. Then if n N, we have that b = b b < b b b Therefore for any > 0, we have found N N such that if n N, then converges to b. Since (a n ) a and b ( ) ( ) b, it follows from part (c) that an bn a b, as desired. <. Hence ( ) Exercise : Abbott Exercise.3. Let x n 0 for all n N. (a) If (x n ) 0, show that ( x n ) 0. (b) If (x n ) x, show that ( x n ) x. Exercise : Abbott Exercise.3.4 Let (a n ) 0, and use the Algebraic Limit Theorem (??) to compute the following limits, and prove your result. (Assume that the sequence (a n ) is such that all fractions below are defined). (a) lim n +a n +3a n 4a n (b) lim n (a n+) 4 a n (c) lim n an +3 an +5 Exercise 3: Abbott Exercise.3.7 Give an example of each of the following, or prove that such a request is impossible. (a) Sequences (x n ) and (y n ), which both diverge, but whose sum (x n + y n ) converges. (b) Sequences (x n ) and (y n ), where (x n ) converges, (y n ) diverges, and (x n + y n ) converges. (c) A convergent sequence ( ) with 0 for all n N such that (/ ) diverges. (d) An unbounded sequence (a n ) and a convergent sequence ( ) with (a n ) bounded. (e) Two sequences (a n ) and ( ), where (a n ) and (a n ) converge but ( ) does not.

4 MAT5 LECTURE 0 NOTES 4 Exercise 4: Abbott Exercise.3.9 (a) Let (a n ) be a bounded (not necessarily convergent) sequence, and assume that lim n = 0. Show that lim n (a n ) = 0. Why are we not allowed to use the Algebraic Limit Theorem (??) to prove this? (b) Can we conclude anything about the convergence of (a n ) if we assume that ( ) converges to some nonzero limit b?. Order Limit Theorem Theorem : Order Limit Theorem (Abbott Theorem.3.4) Let (a n ) and ( ) be sequences of real numbers such that lim n a n = a and lim n = b. Then the following statements hold. (a) If a n 0 for all n N, then a 0. (b) If a n fo all n N, then a b. (c) If there exists c R for which c for all n, then c b. Similarly, if a n c for all n N, then a c. Proof. (a) Suppose, for contradiction, that a < 0. Then let = a, and note that > 0. Since (a n) a, it follows that there exits N N such that if n N, then a n a <. Since N N, we have that a N a < = a N a < a = a N a < a () = a N < a Here () follows by adding a to both sides of the inequality. However a < 0, hence dividing both sides by yields a < 0. Therefore we have that a N < a < 0, contradicting that a n 0 for all n N. (b) Since (a n ) a,?? part (a) implies that ( a n ) a, and therefore?? part (b) implies that ( a n ) b a. Since a n for all n N, it follows that 0 a n for all n N. Therefore applying part (a) of this theorem then yields that 0 b a, which implies that a b, as desired. (c) Let (x n ) be the sequence defined by x n = x for all n N. Then we have that x n for all n. By??, (x n ) c, hence applying part (b) of this proposition yields that c b. A similar argument shows that a c if a n c for all n N. Definition : Eventually 3. Eventuality Let (a n ) be a sequence for which some property of a n holds for all n N for some N N. Then we say that (a n ) eventually has this property.

5 MAT5 LECTURE 0 NOTES 5 Note : Eventuality Properties of the limit of a convergent sequence, in general, do not depend on the behavior of the beginning of the sequence. In this sense, only eventuality matters for the limit. Example : Eventuality (a) Let c R be a real number. A sequence (a n ) is eventually c if there exists some N N such that a n 0 for all n N. (b) Given two sequences (a n ) and ( ), we say that (a n ) is eventually less than or equal to ( ), if there exists some N N such that a n for all n N. Exercise 5: Eventually Show that the Order Limit Theorem (??) still holds if the conditions are relaxed to being only eventually true. Proof. Homework. 4. Monotone Convergence Theorem Definition : Increasing, Decreasing, and Monotone A sequence (a n ) is increasing if a n a n+ for all n N and decreasing if a n a n+ for all n N. A sequence is monotone if it is either increasing or decreasing. Theorem 3: Monotone Convergence Theorem (Abbott Theorem.4.) If a sequence is monotone and bounded, then it converges.