EXAMPLE 5 (a) Approximate the sum of the series 21/n by 15 terms. Estimate the error involved in this approximation. (b terms are required to ensure that the sum is accurate to within

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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(a) Approximate the sum of the series E1/n by
15 terms. Estimate the error involved in this approximation. (b
terms are required to ensure that the sum is accurate to within
EXAMPLE 5
In both parts (a) and (b) we need to know Sf(x)dx
1/x, which satisfies the conditions of the Integral Test, we hav
SOLUTION
1
dr = lim
4x
1
= lim
4t4
4n4
1
4n
4
%D
1
1 1 1
1
E -* S15 =-+
+ -+ ... +
* 1.03692
n=1 n5
15 25 35
155
According to the remainder estimate for the Integral Test, we h
1
R15 < .
-dx =
so the size of the error is at most 0.000004938.
(b) Accuracy to within 0.0005 means that we have to find a val
that Rn < 0.0005. Since
1
R, <
1
dr%3D
4n
4
we want
1
4n
< 0.0005. Solving this inequality, we get
1
n4 >
500
40.0005
n > 4.7287
We need
terms to ensure accuracy to within 0.0005.
Transcribed Image Text:(a) Approximate the sum of the series E1/n by 15 terms. Estimate the error involved in this approximation. (b terms are required to ensure that the sum is accurate to within EXAMPLE 5 In both parts (a) and (b) we need to know Sf(x)dx 1/x, which satisfies the conditions of the Integral Test, we hav SOLUTION 1 dr = lim 4x 1 = lim 4t4 4n4 1 4n 4 %D 1 1 1 1 1 E -* S15 =-+ + -+ ... + * 1.03692 n=1 n5 15 25 35 155 According to the remainder estimate for the Integral Test, we h 1 R15 < . -dx = so the size of the error is at most 0.000004938. (b) Accuracy to within 0.0005 means that we have to find a val that Rn < 0.0005. Since 1 R, < 1 dr%3D 4n 4 we want 1 4n < 0.0005. Solving this inequality, we get 1 n4 > 500 40.0005 n > 4.7287 We need terms to ensure accuracy to within 0.0005.
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