   Chapter 11.3, Problem 41E

Chapter
Section
Textbook Problem

# Show that if we want to approximate the sum of the series ∑ n = 1 ∞ n − 1.001 so that the error is less than 5 in the ninth decimal place, then we need to add more than 1011.301 terms!

To determine

To show: If the sum of the series so that the error is less than 5×109 , then n>1011,301

Explanation

Given:

The series n=1n1.001 and remainder Rn<5×109 .

Result used:

(1) If the function f(x) is continuous, positive and decreasing on [1,) and let an=f(n) , then the series n=1an is convergent if and only if the improper integral 1f(x)dx is convergent.

(2) The function f(x) is decreasing function if f(x)<0.

(3) If the function f(k)=ak , where f is a continuous, positive and decreasing function for xn and an is convergent. If Rn=ssn , then n+1f(x)dxRnnf(x)dx .

Proof:

Consider the function from given series x1.001 .

The derivative of the function is obtained as follows.

f(x)=(1.001)x1.0011=(1.001x2.001)=(1.001x2.001)<0

Since f(x)<0 by the Result (2), the given function is continuous, positive and decreasing on [1,) .

Apply the Result (3) and obtain Rn as follows.

Rnnx1

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