Chapter Sequences Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.2, Slide Section 4.2 Limit Theorems Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.2, Slide To simplify our work with convergent sequences, we prove several useful theorems in this section The first theorem shows that algebraic operations are compatible with taking limits Theorem 4.2.1 Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t Then (a) lim (sn + tn) = s + t (b) lim (ksn) = ks and lim (k + sn) = k + s, for any k ∈ (c) lim (sn tn) = s t (d) lim (sn Proof: (a) For all n ∈ /tn) = s / t, provided that tn ≠ for all n and t ≠ we have ≤ |sn – s| + |tn – t |(sn + tn) – (s + t)| = |(sn – s) + (tn – t)| Given any ε such that |tn – t by the triangle inequality such that |sn – s| < ε /2 for all n ≥ N1 | < ε /2 for all n ≥ N2 Now let N = max {N1, N2} Then for all n ≥ N we have |(sn + tn) – (s + t)| ≤ |sn – s| + |tn – t Thus, lim (sn + tn) = s + t | > 0, since sn → s, there exists N1 ∈ Likewise, there exists N2 ∈ | < ε ε + = ε 2 The proof of (b) is Exercise Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.2, Slide Theorem 4.2.1 Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t Then (c) lim (sn tn) = s t Proof: This time we use the inequality | sn tn – st | = |(sn tn – sn t) + (sn t – st)| ≤ |(sn tn – sn t)| + |(sn t – st)| = | sn | ⋅ | tn – t | + |t | ⋅ |sn – s | We know from Theorem 4.1.13 that the convergent sequence (s ) is bounded n Thus there exists M > such that |Letting s | ≤ M for all n n 1 M = max {M1, | t |}, we obtain the inequality sn tn − st ≤ M tn − t + M sn − s Given any ε > 0, there exist natural numbers N1 and N2 such that < ε /(2M) when n ≥ N and | Now let N = max {N1, N2} Then n ≥ N implies that | tn – t | s – s | < n ε /(2M) when ε sn tn − st ≤ M tn − t + M sn − s < M 2M Thus, lim sn tn= s n ≥ N2 ε ε ε ÷+ M M ÷ = + = ε t Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.2, Slide Theorem 4.2.1 Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t Then (d) lim (sn /tn) = s / t, provided that tn ≠ for all n and t ≠ Proof: Since s That is, given ε n /t = s (1/t ), it suffices from part (c) to show that lim (1/t ) = 1/t n n n n > 0, we must make t − tn 1 − = < ε tn t tn t for all n sufficiently large To get a lower bound on how small the denominator can be, we note that, since t ≠ 0, there exists N ∈ such that n ≥ N implies that | tn – t | < | t | /2 tn = t − (t − tn ) ≥ | t | − t − tn > | t | − There also exists N2 ∈ such that n ≥ N2 implies that |t | |t | = 2 Thus for n ≥ N1 we have by Exercise 3.2.6(a) tn − t < 12 ε | t | Let N = max {N1, N2} Then n ≥ N implies that t − tn 1 t − tn t − tn − = = < < ε tn t tn t t t | tn | |t | Hence lim (1/ tn)= 1/ t ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.2, Slide Example 4.2.2* Show that 5n − lim = 8n − 3n This is Example 4.1.9* from the last section We have sn = 5n − − / n2 = − /n 8n − 3n 2 Now lim (1/n) = 0, so lim (1/n ) = = 0, lim (– /n ) = (– 6)(0) = 0, Likewise, lim (3/n) = 0, And, and lim [5 – (6 /n2)] = so lim [8 – (3/n)] = 5n − lim = 8n − 3n Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.2, Slide Theorem 4.2.4 Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t If sn ≤ tn for all n ∈ , then s ≤ t Then ε = (s – t)/2 > 0, and we have 2ε = s – t and t + ε = s – ε Proof: Suppose that s > t ε ε t Thus there exists N1 ∈ s such that n ≥ N1 implies that s – ε Similarly, there exists N2 ∈ < sn < s + ε such that n ≥ N2 implies that t – ε Let N = max {N1, N2} Then for all n ≥ N we have tn < tn < t + ε < t + ε = s – ε < sn , This contradicts the assumption that sn ≤ tn for all n, and we we conclude that s ≤ t ♦ Corollary 4.2.5 If (tn) converges to t and tn ≥ for all n ∈ , then t ≥ Proof: Exercise 4(b) ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.2, Slide A “ratio test” for convergence Theorem 4.2.7 Suppose that (sn) is a sequence of positive terms and that the sequence of ratios (sn + / sn) converges to L If L < 1, then lim sn = Proof: Corollary 4.2.5 implies L ≥ Suppose L < Then there exists a real number c such that ≤ L < c < Let ε = c – L so that ε > Then since (sn + such that n ≥ N implies that sn +1 − L < ε sn Let k = N + 1 Then for all n ≥ k we have n – ≥ N, so that It follows that, for all n ≥ k, Letting 1/sn) converges to L, there exists N ∈ sn < L + ε = L + (c − L) = c sn −1 < sn < sn−1c < sn−2c < ×××< sk c n−k k M = sk /c , we obtain < sn < M Since 0 < c < 1, Exercise 4.1.7(f ) c n for all n ≥ k implies that lim cn = Thus lim sn = by Theorem 4.1.8 ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.2, Slide To handle a sequence such as (sn) = (n) where the terms get larger and larger, we introduce infinite limits Definition 4.2.9 A sequence (sn) is said to diverge to + ∞, and we write lim sn = + ∞ if for every M ∈ there exists a natural number N such that n ≥ N implies that sn > M A sequence (sn) is said to diverge to – ∞, and we write lim sn = – ∞ if for every M ∈ there exists a natural number N such that n ≥ N implies that sn < M Example 4.2.11 Show that 4n − lim = + ∞ n+2 We find that 4n – This time we need a lower bound on the numerator 2 ≥ 4n – n = 3n , when n ≥ For an upper bound on the denominator, we have n + Thus for n ≥ we obtain 4n − 3n 3n ≥ = n+2 2n So given any M ∈ ≤ n + n = 2n, when n ≥ To make 3n/2 > M, we want n > 2M /3 , take N > max {2, 2M / 3} The formal proof is in the text Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.2, Slide We conclude with two theorems for infinite limits The proofs are left for the exercises Theorem 4.2.12 Suppose that (sn) and (tn) are sequences such that sn ≤ tn for all n ∈ (a) If lim sn = + ∞, then lim tn = + ∞ (b) If lim tn = – ∞, then lim sn = – ∞ Theorem 4.2.13 Let (sn) be a sequence of positive numbers Then lim sn = + ∞ Copyright © 2013, 2005, 2001 Pearson Education, Inc if and only if lim (1 / sn) = Section 4.2, Slide 10 ... (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t Then (a) lim (sn + tn) = s + t (b) lim (ksn) = ks and lim (k + sn) = k + s, for any k ∈ (c) lim (sn tn) = s t (d) lim (sn Proof: ... (sn) is said to diverge to + ∞, and we write lim sn = + ∞ if for every M ∈ there exists a natural number N such that n ≥ N implies that sn > M A sequence (sn) is said to diverge to – ∞, and we write... numerator 2 ≥ 4n – n = 3n , when n ≥ For an upper bound on the denominator, we have n + Thus for n ≥ we obtain 4n − 3n 3n ≥ = n+2 2n So given any M ∈ ≤ n + n = 2n, when n ≥ To make 3n/2 > M, we want