Analysis with an introduction to proof 5th by steven lay ch08

Chapter Infinite Series Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.1, Slide Section 8.1 Convergence of Infinite Series Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.1, Slide We begin with some notation Let (an) be a sequence of real numbers We use the notation n ∑ k =m ak ∑ or n a k =m k to denote the sum am + am  + 1 + … + a , where n ≥ m n Using (an), we can define a new sequence (sn) of partial sums given by sn = n ∑ ak k =1 = a1 + a2 + L + an ∞ ∑ an We also refer to the sequence (sn) of partial sums as the infinite series If (sn) converges to a real number s, we say that the series and we write ∑ We also refer to s as the sum of the series A series that is not convergent is divergent If lim sn = + ∞, and we write  ∞, ∞ a n =1 n ∑ = s ∞ ∑ n =1 an ∞ ∑ n =1 an ∞ an = + ∞ we say that∑the n =1 series Copyright © 2013, 2005, 2001 Pearson Education, Inc ∞ a is convergent n =1 n n =1 diverges to + Section 8.1, Slide As defined above, the symbol ∑ ∞ an in two ways: n =1is used It is used to denote the sequence (sn) of partial sums It is used to denote the limit of the sequence (sn) of partial sums, provided that this limit exists The context will make the intended meaning clear ∞ Remember that an n =1 ∑ represents a limit of partial sums In particular, an expression such as a1 + a2 + a3 + … + an + …, really means… lim ( a1 + a2 + L + an ) n →∞ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.1, Slide Example 8.1.1 ∞ , we have ∑ then(partial sums given by n + 1) n =1 For the infinite series sn = 1 1 + + +L + ×2 ×3 ×4 n( n + 1)  1   1   1  1 =  − ÷+  − ÷+  − ÷ + L +  − ÷ 1       n n +1 This is a “telescoping” series because of the way the terms in the partial sums cancel sn = − n +1 → 1, as n → ∞ ∞ So we write ∑ n(n + 1) = 1, n =1 Copyright © 2013, 2005, 2001 Pearson Education, Inc or 1 1 + + + + + L = 12 20 30 Section 8.1, Slide Example 8.1.2 ∞ ∑ 1n The harmonic series has the partial sums n =1 sn = + 1 1 + + +L + n In Example 4.3.13 we saw that the sequence (s ) is divergent Thus the harmonic series n is also divergent Since the partial sums form an increasing sequence, we must have lim s = + ∞, so ∞ ∑ 1n n = + ∞ n =1 Theorem 8.1.4 Suppose that Σ an = s and Σ bn = Σ  (ka ) = ks, for every k ∈ n t with s, t ∈ Then Σ (an + bn) = s + t and The proof follows from the corresponding result for sequences Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.1, Slide Theorem 8.1.5 If Σ an is a convergent series, then lim an = Proof : If Σ  a converges, then the sequence (s ) of partial sums must have a finite limit, say s n n But a = s – s – 1, so n n n lim a n = lim s – lim s = s – s = ♦ n n–1 Note that the converse of Theorem is not true ∞ ∑ 1n The harmonic series is divergent, even though lim 1/n = n =1 Corresponding to the Cauchy criterion for sequences (Theorem 4.3.12), we have the following result for series Theorem 8.1.6 (Cauchy Criterion for Series) The infinite series Σ  a converges iff for each ε n then > there exists a natural number N such that, if n ≥ m ≥ N, | a + a     + … + a | < ε m m+1 n     Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.1, Slide Example 8.1.7 ∞ One of the most useful series is the geometric series ∑ rn n=0 − r n+1 = 1+ r + r +L + r = In Exercise 3.1.7 we saw that for r ≠ we have sn = ∑ r 1− r k =0 for all n ∈ n k n But Exercise 4.1.7(f ) shows that lim r n + 1 = when | r | < 1, so we conclude that ∞ n r ∑ = lim sn = n=0 , 1− r for | r | < If | r | ≥ 1, then the sequence (r n) does not converge to zero, and it follows from Theorem 8.1.5 that the series Σ r n diverges Here is a geometric view of the geometric series: Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.1, Slide The Geometric Series Let A = (0,0) and B = (1,0) C Draw a line of slope r through A, with < r < Draw a line of slope through B Let C be the point where they meet Let P1 be the point on AC directly above B How long is BP1? P1 Slope is r r A B Slope is s = ∞ ∑ rn n=0 Copyright © 2013, 2005, 2001 Pearson Education, Inc = lim sn = + r + r + r + r + L n→∞ Section 8.1, Slide The Geometric Series Let P2 be the point on BC at the same height as P1 C How long is P1 P2? Let P3 be the point on AC directly above P2 r P5 How long is P2 P3? r P3 Repeat this process r P4 r P1 r P2 Slope is r r A B Slope is s = ∞ ∑ rn n=0 Copyright © 2013, 2005, 2001 Pearson Education, Inc = lim sn = + r + r + r + r + L n→∞ Section 8.1, Slide 10 The Geometric Series Now project the horizontal lengths down onto the x-axis C And project down the point C as well, and call it D Look at the partial sums: s0 = s2 = + r + r 2 s3 = + r + r + r r P3 r etc And r P5 s1 = + r P4 r lim sn = s n→∞ P1 r P2 r A r B s0 s = s1 ∞ ∑ rn n=0 Copyright © 2013, 2005, 2001 Pearson Education, Inc r r s2 r s3 D s4 s = lim sn = + r + r + r + r + L n→∞ Section 8.1, Slide 11 The Geometric Series How long is CD? It’s the same as BD C Look at the similar triangles ∆ABP1 and ∆ADC The corresponding sides are proportional P5 s = s–1 r r P3 r P4 r P1 s–1 r P2 r A r B s0 s = s1 ∞ ∑ rn n=0 Copyright © 2013, 2005, 2001 Pearson Education, Inc r r s2 r s3 D s4 s = lim sn = + r + r + r + r + L n→∞ Section 8.1, Slide 12 The Geometric Series Solve this s = s–1 for s r C sr = s – P5 s – sr = s(1 – r) = s = r P3 r P4 r 1–r P1 s–1 r P2 r A r B s0 s = s1 ∞ ∑ rn n=0 Copyright © 2013, 2005, 2001 Pearson Education, Inc r r s2 r s3 D s4 s = lim sn = + r + r + r + r + L n→∞ Section 8.1, Slide 13 The Alternating Geometric Series Slope is – r s = ∞ ∑ (−1) n n Slope is r =1− r + r − r + r − r + L n=0 B A r P3 P4 r r r P6 P2 Copyright © 2013, 2005, 2001 Pearson Education, Inc r P5 r P1 Section 8.1, Slide 14 The Alternating Geometric Series Slope is – r s = ∞ ∑ (−1) n n Slope is r =1− r + r − r + r − r + L n=0 B A r P3 P4 r r r r r4 r P6 P2 Copyright © 2013, 2005, 2001 Pearson Education, Inc r P5 r P1 Section 8.1, Slide 15 The Alternating Geometric Series Slope is – r s = ∞ ∑ (−1) n n r =1− r + r − r + r − r + L n=0 s1 Slope is s3 s5 s4 s s0 s2 D A r B P3 P4 Look at the partial sums: r r s0 = r s2 = – r + r r C r4 s1 = – r r 2 P6 P2 Copyright © 2013, 2005, 2001 Pearson Education, Inc r P5 r P1 Section 8.1, Slide 16 The Alternating Geometric Series Slope is – r s = ∞ ∑ (−1) n n r =1− r + r − r + r − r + L n=0 s1 Slope is s3 s5 s4 s s0 s2 D A B P3 P4 Similar triangles give us r s = 1–s r r or s = C r4 r r P6 1+r P2 Copyright © 2013, 2005, 2001 Pearson Education, Inc r P5 r P1 Section 8.1, Slide 17 Practice 8.1.8 n +1 2 + + +L + n +L Find the sum of the series 2n+1 + + +L + n +L = We have A geometric series with r = 2/3 n  ∞  n  2 ∑  ÷ =  ∑  ÷ ÷÷ n=0  n=0  ∞   = 2  1− ÷ ÷  3 = 2(3) = n ∞ 2 ? ∑  ÷ n =1 And, what is This series is the same as the first series, except without the first term ∞ So, n 2 ∑  ÷ = − = n =1 Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 8.1, Slide 18 ... given by sn = n ∑ ak k =1 = a1 + a2 + L + an ∞ ∑ an We also refer to the sequence (sn) of partial sums as the infinite series If (sn) converges to a real number s, we say that the series and... intended meaning clear ∞ Remember that an n =1 ∑ represents a limit of partial sums In particular, an expression such as a1 + a2 + a3 + … + an + …, really means… lim ( a1 + a2 + L + an ) n →∞... write ∑ We also refer to s as the sum of the series A series that is not convergent is divergent If lim sn = + ∞, and we write  ∞, ∞ a n =1 n ∑ = s ∞ ∑ n =1 an ∞ ∑ n =1 an ∞ an = + ∞ we say that∑the