Chapter Sequences Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide Section 4.1 Convergence Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide A sequence S is a function whose domain is the set of natural numbers We usually write sn instead of S(n) We may describe the sequence S as (sn) or by listing the elements (s1, s2, s3, …) Sometimes we just give a formula for the typical nth term: ( 1n ) = ( 1, , , , ) Sometimes we want to change the domain to include or start at something other than ∞ ( sn )∞ or ( s ) n=0 n n=m Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide Example 4.1.1 n Consider the sequence (sn ) given by sn = + (–1) Here are the first few terms: s1 = + (–1) = s2 = + (–1) = + = s3 = + (–1) = – = s4 = + (–1) = + = 2, and so on So, (sn) = (0, 2, 0, 2, 0, 2, …) Note that the terms of a sequence not have to be distinct We consider s2 and s4 to be different terms, even though their values are both The range of this sequence is the set of values obtained: {0, 2} So the range of a sequence may be finite, even though the sequence will always have infinitely many terms Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide Sometimes the values in a sequence get “close” to a fixed number Like (1/n) gets close to Definition 4.1.2 A sequence (sn) is said to converge to the real number s provided that for every ε > there exists a natural number N such that for all n ∈ , n ≥ N implies that |sn – s| < ε If (sn) converges to s, then s is called the limit of the sequence (sn) We write limn→ ∞ sn = s, lim sn = s, or sn → s If a sequence does not converge to a real number, it is said to diverge Note the order of the quantifiers in the definition The number N may depend on ε We don’t have to find one N that works for all ε Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide Example 4.1.3 ( 1n ) = ( 1, Consider the sequence , , , , ) If we were to graph this sequence as a function it would look like this: (1, 1) • (2, 1/2) ε • (3, 1/3) • = 0.3 It appears that lim sn = (4, 1/4) • (5, 1/5) • • … Let’s show that this is true If, for example, ε = 0.3, how should N be chosen so that n ≥ N implies |1/n – 0| < ε ? We have |1/n – 0| = 1/n, and 1/n is less than 0.3 when n ≥ So choose any natural number N such that N ≥ This is an illustration, but not a proof We must show that for every ε > there exists an N that works Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide Example 4.1.3 ( 1n ) = ( 1, Consider the sequence , , , , ) If we were to graph this sequence as a function it would look like this: (1, 1) • (2, 1/2) • (3, 1/3) It appears that lim sn = (4, 1/4) • • (5, 1/5) • • … Let’s show that this is true Given any ε > 0, the Archimedean property says there exists N ∈ such that < 1/N < ε Thus for any n ≥ N we have n Copyright © 2013, 2005, 2001 Pearson Education, Inc – = n ≤ N < ε Section 4.1, Slide Example 4.1.6* n + 5n = n –7 Show that lim *Similar to 4.1.6, but with different numbers n + 5n Given any ε > 0, we want to make –0 < ε n –7 By making n ≥ we can remove the absolute value signs since n – will be positive n + 5n So we want to know how large n has to be in order for n –7 < ε This is hard to solve for n, so we seek an estimate of how large the left side will be That is, we seek an upper bound for the numerator and a lower bound for the denominator 2 For large values of n, the numerator behaves like n , so we want n + 5n ≤ bn 3 And the denominator behaves like n , so we want n – ≥ cn Then we will have n + 5n n –7 ≤ = bn cn b c n and the latter expression is relatively easy to make small Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide Example 4.1.6* n + 5n = n –7 Show that lim n + 5n Given any ε > 0, we want to make n + 5n ≤ n –7 With n ≥ we want bn cn 2 Let b = 2, so that n + 5n ≤ 2n We have n ≥ 5n Now let c = n Any b > would work Why b = 2? and n ≥ , so that n – ≥ We have –0 < ε n –7 n ≥ 14, n ≥ 7, n + 5n n –7 Copyright © 2013, 2005, 2001 Pearson Education, Inc ≤ < c < would work In that case we have –0 Any c with and n ≥ if n ≥ All three conditions will be satisfied ? Why c = 2n (1/2)n = n Section 4.1, Slide Example 4.1.6* Show that lim n + 5n = n –7 n + 5n Given any ε > 0, we want to make n + 5n When n ≥ we have To make n < ε, we need n > –0 < ε n –7 –0 n –7 ≤ n ε There are two conditions to be satisfied We can accomplish both by letting N be a natural number such that N > max {5, }.4 This is possible because of the Archimedean property ε So when n ≥ N, both conditions are satisfied Now let’s organize this into a proof Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide 10 Example 4.1.6* n + 5n = n –7 Show that lim Given any ε > 0, take N ∈ such that N > max Then for n ≥ N we have n > and n > {5, }.4 ε ε Since n > we have n + 5n ≤ 2n and n – ≥ (1/2)n 2 Thus for n ≥ N we have n + 5n n –7 –0 = n + 5n n –7 ≤ 2n (1/2)n = n < ε ♦ This is a lot of work, but it can be reduced somewhat by the following theorem Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide 11 Theorem 4.1.8 Let (sn) and (an) be sequences of real numbers and let s ∈ and some m ∈ If for some k > we have |sn – s| ≤ k | |, an for all n ≥ m, and if lim an = 0, then it follows that lim sn = s Proof: Given any ε implies that | an | < ε /k > 0, since lim an = there exists N1 ∈ such that n ≥ N1 Now let N = max {m, N1} Then for n ≥ N we have n ≥ m and n ≥ N1, so that ε  sn − s ≤ k an < k  ÷ = ε k Thus lim sn = s ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide 12 Example 4.1.9* Show that lim 5n – = 8n – 3n To apply the theorem we need to find an upper bound for 5n – 8n – 3n – = 15n – 48 when n is sufficiently large 8(8n – 3n) The numerator is easy, since |15n – 48| < 15n for all n ≥ 2 2 For the denominator, we have 8n – 3n ≥ 7n when n ≥ 3n and n ≥ Thus when n ≥ 3, we have 5n – 8n – 3n – = 15n – 48 8(8n – 3n) Since lim (1/n) = 0, Theorem 4.1.8 implies that Copyright © 2013, 2005, 2001 Pearson Education, Inc < 15n 8(7n ) = 15 56 n lim 5n – 8n – 3n = Section 4.1, Slide 13 Note: A sequence (sn) is bounded if the range {sn : n ∈ } is a bounded set That is, if there exists an M > such that | sn | ≤ M for all n ∈ Theorem 4.1.13 Every convergent sequence is bounded Proof: Let (sn) be a convergent sequence and let lim sn = s convergence with ε = 1, we obtain N ∈ From the definition of such that | sn – s | < whenever n ≥ N   Thus for n ≥ N we have | sn | < | s | + by Exercise 3.2.6(b)         If we let   M = max {| s1 |, | s2 |, …, | sN |, | s | +1},                 then we have | sn | ≤ M for all n ∈ , so (sn) is bounded ♦     Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide 14 Theorem 4.1.14 If a sequence converges, its limit is unique Proof: Suppose sn → s and sn → t there exists N1 ∈ Given any ε > 0, such that |sn – s| < And there exists N2 ∈ ε , for every n ≥ N1 ε , for every n ≥ N2 such that |sn – t | < Therefore, if n ≥ max {N1, N2} then from the triangle inequality we have ≤ |s – sn | + |sn – t | |s – t | = |s – sn + sn – t | < ε + ε = ε Since this holds for all ε > 0, we must have s = t ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 4.1, Slide 15 ... *Similar to 4.1.6, but with different numbers n + 5n Given any ε > 0, we want to make –0 < ε n –7 By making n ≥ we can remove the absolute value signs since n – will be positive n + 5n So we want to. .. ∈ and some m ∈ If for some k > we have |sn – s| ≤ k | |, an for all n ≥ m, and if lim an = 0, then it follows that lim sn = s Proof: Given any ε implies that | an | < ε /k > 0, since lim an. .. denominator 2 For large values of n, the numerator behaves like n , so we want n + 5n ≤ bn 3 And the denominator behaves like n , so we want n – ≥ cn Then we will have n + 5n n –7 ≤ = bn cn b c n and