University:
Miami University
Course:
MTH 151 | Calculus I
Academic year:

2020
Views:

194

Pages:

9
Author:

ilikemyicons0y5

Approximating the Sum of a Convergent Series The Calculus BC Course Description mentions how technology can be used to explore convergence and divergence of series, and lists various tests for convergence and divergence as topics to be covered. But no specific mention is made of actually estimating the sum of a series, and the only discussion of error bounds is for alternating series and the Lagrange error bound for Taylor polynomials. With just a little additional effort, however, students can easily approximate the sums of many common convergent series and determine how precise those approximations will be. Approximating the Sum of a Positive Series Here are two methods for estimating the sum of a positive series whose convergence has been established by the integral test or the ratio test. Some fairly weak additional requirements are made on the terms of the series. Proofs are given in the appendix. ∞ n n =1 k =1 Let S = ∑ an and let the nth partial sum be Sn = ∑ ak . 1. Suppose an = f (n) where the graph of f is positive, decreasing, and concave up, and the improper integral Sn + ∫ ∞ n +1 f ( x) dx + ∞ ∫1 f ( x) dx converges. Then ∞ an+1 a < S < Sn + ∫ f ( x) dx − n+1 . n 2 2 In Example 2 you will see how this holds for n ≥ N if we only know f is positive, decreasing, and concave up on the interval [N, ∞). (1) Special Focus: Calculus a 2. Suppose (an ) is a positive decreasing sequence and lim n+1 = L < 1. n→∞ an an+1  L  an+1 If decreases to the limit L , then S n + an  + , (2) < S < S n  an  1 − L  an+1  1 − a   n  If an+1  L  an+1 increases to the limit L , then S n + < S < S n + an  .  an  1 − L  an+1  1 − a   n  ∞ Example 1: S = ∑ 1 n =1 n The function f ( x) = (3) 2 1 is positive with a graph that is decreasing and concave up x2 ∞ f (x )dx converges. This series for x ≥ 1, and an = f (n) for all n . In addition, ∫ 1 converges by the integral test. By inequality (1), 1 1 1 1 Sn + + < S < Sn + − . 2 n + 1 2(n + 1) n 2(n + 1) 2 This inequality implies that S is contained in an interval of width 1 2 1 1 − − = . 2 n 2(n + 1) n + 1 n(n + 1) 2 If we wanted to estimate S with error less than 0.0001, we could use a value of n 1 with 2 < 0.0002 and then take the average of the two endpoints in inequality n (n + 1) (4) as an approximation for S. The table feature on a graphing calculator shows that n € = 17 is the first value of n that works. Inequality (4) then implies that 1.6449055 < S < 1.64508711 and a reasonable approximation would be S≈ 1.6449055 + 1.6450870 ≈ 1.645 2 to three decimal places. With n = 100, inequality (4) actually shows that 1.6449339 < S < 1.6449349, and hence we know for sure that S = 1.64493.... Of course, in this case we actually know that SS == π2 6 ==11.644934066.... .644934066... Notice also that S100 ≈ 1.6349839, so the partial sum with 100 terms is a poor approximation by itself. 1. We will use the convention for positive endpoints of truncating the left endpoint of the interval and rounding up the right endpoint. This will make the interval slightly larger than that given by the actual symbolic inequality. Approximating the Sum of a Convergent Series ∞ Example 2: S = ∑ x n =1 n n 4 +1 . The graph of f is decreasing and concave up for x ≥ 2. Also x +1 ∞ ∞ n x π 1 2 and so the improper integral converges. Let T = = − arctan( n ) ∑ n4 + 1 ∫ n x4 + 1 4 2 n=2 € with partial sums Tn for n ≥ 2. By inequality (1), Let f ( x) = Tn + 4 π 1 n +1 π 1 n +1 2 − arctan((n + 1) 2 ) + < T < T + − arctan( n ) − n 4 2 4 2 2((n + 1) 4 + 1) 2((n + 1) 4 + 1) for n ≥ 2. But now add a1 , the first term of the series for S, to each term of this inequality to see that π 1 n +1 π 1 n +1 Sn + − arctan((n + 1) 2 ) + an+1 + (an+1 − an+ 2 ). and so ∫ n 2 Hence ∞ 1 1 1 … ∫ n f ( x) dx > an+1 + 2 (an+1 − an+2 )+ an+2 + 2 (an+2 − an+3 )+ an+3 + 2 (an+3 − an+4 )+ L 1 … = an+1 + an+1 + an+ 2 + an+3 + L 2 1 = an+1 + S − Sn 2 Proof of Inequalities (2) and (3) ∞ n n =1 k =1 Let S = ∑ an and let Sn = ∑ ak . Suppose (an ) is a positive decreasing sequence and a lim n+1 = L < 1 , where the ratios decrease to L . The series converges by the ratio n→∞ an test. a a Let r = n+1 < 1. Then k +1 < r for all k ≥ n . Hence an ak an+1 < an r an+ 2 < an+1r < an r 2 an+3 ∑ an Lk = an n L . 1− L Combining these two results gives inequality (2). A similar argument for the inequalities with r and L reversed proves inequality (3). Proof of Inequalities (6) and (7) ∞ n n =1 k =1 Let S = ∑ (−1) n+1 an and let Sn = ∑ (−1) k +1 ak , where (an ) is positive decreasing sequence that converges to 0. Let bn = an − an+1 , where we assume that the sequence …) and ( bn ) also decreases monotonically to 0. Then S = Sn + (−1) n (bn+1 + bn+3 + bn+5 +L Appendix …). S = Sn−1 + (−1) n+1 (bn + bn+ 2 + bn+ 4 + L Because the sequence ( bn ) decreases, … < bn + bn+ 2 + bn+ 4 + L … = S − Sn−1 . S − Sn = bn+1 + bn+3 + bn+5 + L Therefore S − Sn < S − Sn−1 . Similarly, S − Sn+1 < S − Sn . But S lies between the successive partial sums, so it follows that an = Sn − S n−1 = S − S n + S − S n−1 > 2 S − S n and an+1 = Sn+1 − Sn = S − Sn+1 + S − S n < 2 S − S n . Combining these two results shows that inequalities (6) and (7) can be obtained. an+1 a < S − Sn < n , from which 2 2

Approximating the Sum of a Convergent Series

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