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Cse547, Math547 DISCRETE MATHEMATICS cse547, math547 DISCRETE MATHEMATICS Professor Anita Wasilewska LECTURE 10 CHAPTER 2 SUMS Part 1: Introduction - Lecture 5 Part 2: Sums and Recurrences (1) - Lecture 5 Part 2: Sums and Recurrences (2) - Lecture 6 Part 3: Multiple Sums (1) - Lecture 7 Part 3: Multiple Sums (2) - Lecture 8 Part 3: Multiple Sums (3) General Methods - Lecture 8a Part 4: Finite and Infinite Calculus (1) - Lecture 9a Part 4: Finite and Infinite Calculus (2) - Lecture 9b Part 5: Infinite Sums- Infinite Series - Lecture 10 CHAPTER 2 SUMS Part 5: Infinite Sums- Infinite Series - Lecture 10 Infinite Sums (Series) n We extend now the notion of a finite sum Σk=1 ak to an infinite sum 1 Σn=1 an For a given a sequence fangn2N−{0g, i.e the sequence a1;a2;a3;::::an;:: we consider a following (infinite) sequence n n+1 S1 = a1; ::::; Sn = Σk=1 ak ; Sn+1 = Σk=1 ak ; ::::::: and define the infinite sum as follows Infinite Sum Definition Definition 1 n If the limit of the sequence fSn = Σk=1 ak gn2N−{0g exists we call it an infinite sum of the sequence fangn2N−{0g We write it as 1 n Σ an = lim Sn = lim Σ ak n=1 n!1 n!1 k=1 The sequence fSngn2N−{0g is called its sequence of partial sums Infinite Sum Definition Definition 2 If the limit limn!1 Sn exists and is finite, i.e. lim Sn = S n!1 1 then we say that the infinite sum Σn=1 an converges to S and we write 1 n Σ an = lim Σ ak = S n=1 n!1 k=1 otherwise the infinite sum diverges Observation Observation 1 In a case when all elements of the sequence fangn2N−{0g are equal0 starting from a certain k ≥ 1 1 the infinite sum Σn=1 an becomes a finite sum The infinite sum is a generalization of the finite one, and this is why we keep the similar notation Example 1 Example 1 k The infinite sum of a geometric sequence an = x for x ≥ 0, i.e. the sum 1 n Σn=1 x converges if and only if j x j< 1 It is true because n k x−xn+1 x(1−xn) Σk=1 x = Sn = 1−x = 1−x and x(1 − xn) x x lim = lim (1 − xn) = iff j x j< 1 n!1 1 − x n!1 1 − x 1 − x Moreover x Σ1 xk = n=1 1 − x More Examples Example 2 1 The series Σn=1 1 diverges to 1 as n Sn = Σk=11 = n and lim Sn = lim n = 1 n!1 n!1 More Examples Example 3 1 n The infinite sum Σn=0 (−1) diverges Proof We use the Perturbation Method + = + n Sn an+1 a0 ∑k=0ak+1 to eveluate 1 + (−1)n 1 (−1)n S = Σn (−1)k = = + n k=0 2 2 2 and we prove that 1 (−1)n lim ( + ) does not exist n!1 2 2 More Examples Example 4 1 1 The infinite sum Σn=0 (k+1)(k+2) converges to1; i.e. 1 Σ1 = 1 n=0 (k + 1)(k + 2) n 1 Proof: first we evaluate Sn = Σk=0 (k+1)(k+2) as follows 1 S = Σn = Σn k −2 =Σ n+1k −2 dk n k=0 (k + 1)(k + 2) k=0 k=0 1 n+1 1 = − = − + 1 k + 1 0 n + 2 and 1 lim Sn = lim − + 1 = 1 n!1 n!1 n + 2 Definition Definition 3 For any infinite sum (series) 1 Σn=1an a sum (series) 1 rn = Σm=n+1 am is called its n-th remainder Fact 1 Fact 1 1 If the infinite sum Σn=1an converges, 1 then so does its n-th remainder rn = Σm=n+1 am Proof: 1 Assume that Σn=1an converges n Let’s denote Sn = Σm=1am and we have that 1 S = limn!1 Sn = Σm=1am n Observe that rn = S − Σm=1am = S − Sn By definition, rn converges iff limn!1 rn exists and is finite. We evaluate limn!1 rn = S − limn!1 Sn = S − S = 0 what ends the proof General Properties of Infinite Sums Theorem 1 Theorem 1 If the infinite sum 1 Σ an converges; then lim an = 0 n=1 n!1 Proof: observe that an = Sn − Sn−1 and hence lim an = lim Sn − lim Sn−1 = 0 n!1 n!1 n!1 as limn!1 Sn = limn!1 Sn−1 Theorem 1 Remark 1 The reverse statement to the Theorem 1, namely a statement 1 If lim an = 0 then Σ an converges n!1 n=1 is not always true as there are infinite sums with the term converging to zero that are not convergent Observe that Theorem 1 can be re-written as follows Theorem 1 1 If lim an , 0 then Σ an diverges n!1 n=1 Example 5 Example 5 1 1 The infinite harmonic sum H = Σn=1 n DIVERGES to 1, even if its -th term converges to 0, i.e. 1 1 1 Σn=1 n = 1 and limn!1 n = 0 The infinite harmonic sum provides an example of an 1 infinite diverging sum Σn=1an, such that limn!1 an = 0 Properties Definition 4 Infinite sum 1 Σn=1an is bounded if its sequence of partial sums n Sn = Σk=1 ak is bounded; i.e. there is a number M 2 R such that Sn < M, for all n 2 N Fact 2 Every convergent infinite sum is bounded Properties Theorem 2 If the infinite sums 1 1 Σn=1an; Σn=1bn converge then the following properties hold. 1 1 1 Σn=1(an + bn) = Σn=1an + Σn=1bn; and 1 1 Σn=1can = cΣn=1an; c 2 R Alternating Infinite Sums Definition Definition 5 An infinite sum 1 n+1 Σn=1(−1) an; for an ≥ 0 is called alternating infinite sum (alternating series) Example 6 Consider 1 n+1 Σn=1(−1) = 1 − 1 + 1 − 1 + : If we group the terms in pairs, we get (1 − 1) + (1 − 1) + :::: = 0 but if we start the pairing one step later, we get 1 − (1 − 1) − (1 − 1) − ::::: = 1 − 0 − 0 − 0 − ::: = 1 Alternating Series The Example 6 shows that grouping terms in a case of infinite sum can lead to inconsistencies (contrary to the finite case) Look also example on page 59 of our BOOK We need to develop some strict criteria for manipulations and convergence/divergence of alternating series Alternating Series Theorem Theorem 3 Given an alternating infinite sum 1 n+1 Σn=1(−1) an such that 1. an ≥ 0, for all n 2. sequence fang is decreasing. i.e. a1 ≥ a2 ≥ a3 ≥ :::: 3. limn!1 an = 0 1 n+1 Then the sum Σn=1(−1) an converges, i.e. 1 n+1 Σn=1(−1) an = S n k+1 Moreover the partial sums Sn = Σk=1(−1) ak fulfill the condition 1 n+1 S2n ≤ Σn=1(−1) an ≤ S2n+1 for all n 2 N+ Alternating Series Theorem Proof Proof Evaluate 2n+2 k+1 S2(n+1) = S2n+2 = Σk=1 (−1) ak 2n k+1 2n+2 2n+3 = Σk=1(−1) ak + (−1) a2n+1 + (−1) a2n+2 = S2n + (a2n+1 − a2n+2) By 2. we know that sequence fang is decreasing hence a2n+1 − a2n+2 ≥ 0 and so S2n+2 ≥ S2n i.e we proved that the sequence of S2n is increasing Alternating Series Theorem Proof We are going to prove now that the sequence of S2n is also bounded Observe that 2n+1 S2n = a1 − a2 + a3 − a4 + + (−1) a2n = a1 − (a2 − a3) − (a4 − a5) + ::: − a2n By 2. ak − ak+1 ≥ 0 for k = 2;3;:::;2(n − 1) and by 1. a2n ≥ 0, so −a2n ≤ 0 and we get that S2n ≤ a1 what proves that S2n is bounded Alternating Series Theorem Proof We know that any bounded and increasing sequence is is convergent, so we proved that S2n converges Let denote limn!1 S2n = g To prove that 1 n+1 Σ (−1) an = lim Sn n=1 n!1 converges we have to show now that also lim S2n+1 = g n!1 Observe that S2n+1 = S2n + a2n+1 and we get lim S2n+1 = lim S2n + lim a2n+1 = g n!1 n!1 n!1 as we assumed in 3. that limn!1 an = 0 Alternating Series Theorem Proof We proved that the sequence S2n is creasing We prove, in a similar way (exercise!) that the sequence fS2n+1g is decreasing Hence 1 n+1 S2n ≤ lim S2n = g = Σ (−1) an n!1 n=1 and 1 n+1 S2n+1 ≥ lim S2n+1 = g = Σ (−1) an n!1 n=1 what means that 1 n+1 S2n ≤ Σn=1(−1) an ≤ S2n+1 It ends the proof of the Theorem 3 Example Example 7 Consider the ANHARMONIC series (infinite sum) 1 1 1 1 AH = Σ1 (−1)n+1 = 1 − + − :: n=1 n 2 3 4 1 1 1 Observe that an = n ≥ 0, n ≥ n+1 i.e. an ≥ an+1, for all n, and limn!1 an = 0 So the assumptions of the Theorem 3 are fulfilled forAH and henceAH converges In fact, it is proved (by analytical methods, not ours) that 1 AH = Σ1 (−1)n+1 = ln2 n=1 n Example A series (infinite sum) 1 1 1 1 1 Σ1 (−1)n = 1 − + − + n=0 2n + 1 3 5 7 9 converges by Theorem 3 Proof is similar to the one in the Example 7 It also is proved (by analytical methods, not ours) that 1 p Σ1 (−1)n = n=0 2n + 1 4 and hence we have that 4 p = Σ1 (−1)n n=0 2n + 1 Generalization of Theorem 3 Theorem 4 ABEL Theorem IF a sequence fang fulfils the assumptions of the Theorem 3 i.e.
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