For an alternating series whose summands are decreasing in magnitude, the true sum S lies between any two successive partial sums: (*) min {SN, SN+1} <≤ S≤ max {SN, SN+1}. Consider S=> TUNC1 (-1)+1 1 n5 and write Answer: S (a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10-⁹. SN=(-1)+1 n5 n=1 Answer: Nmin == (b) Using the N found in part (a), approximate S by the midpoint of the interval implicit in line (*). A spreadsheet may be helpful to calculate the sum SN. SN + SN+1 2

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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For an alternating series whose summands are decreasing in magnitude, the true sum S lies between any two successive partial sums:
(*)
min {SN, SN+1} ≤ S≤ max {SN, SN+1}.
Consider S =
n=1
(-1)+1
n5
and write
Answer: S
SN =
N (−1)n+1
n5
(a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10-⁹.
SN + SN+1
2
n=1
Answer: Nmin =
(b) Using the N found in part (a), approximate S by the midpoint of the interval implicit in line (*). A spreadsheet may be helpful to calculate the sum SN.
Transcribed Image Text:For an alternating series whose summands are decreasing in magnitude, the true sum S lies between any two successive partial sums: (*) min {SN, SN+1} ≤ S≤ max {SN, SN+1}. Consider S = n=1 (-1)+1 n5 and write Answer: S SN = N (−1)n+1 n5 (a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10-⁹. SN + SN+1 2 n=1 Answer: Nmin = (b) Using the N found in part (a), approximate S by the midpoint of the interval implicit in line (*). A spreadsheet may be helpful to calculate the sum SN.
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