   Chapter 11.3, Problem 13E

Chapter
Section
Textbook Problem

# Determine whether the series is convergent or divergent.13. 1 3 + 1 7 + 1 11 + 1 15 + 1 19 + ⋯

To determine

Whether the series is convergent or divergent.

Explanation

Given:

The series is 13+17+111+115+119+... .

Result used:

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

Chain rule: d[f(x)]ndx=n[f(x)]n1f(x)

Definition used:

The improper integral is divergent if the limit does not exist.

Calculation:

The given series can be expressed as follows,

13+17+111+115+119+...=1(4(1)1)+1(4(2)1)+1(4(3)1)+1(4(4)1)+1(4(5)1)+...=n=11(4n1)=n=1(4n1)1

Consider the function from given series (4x1)1 .

f(x)=(1)(4x1)11[d(4x1)dx]     (Chain rule)=[(4x1)2(4)]=[4(4x1)2]

Since f(x)<0 , the given function is decreasing by using the Result (2).

Clearly, the function f(x) is continuous, positive and decreasing on [1,)

Use the result stated above, the series is convergent if the improper integral 11(4x1)dx is convergent

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