   Chapter 11.3, Problem 6E

Chapter
Section
Textbook Problem

# Use the Integral Test to determine whether the series is convergent or divergent.6. ∑ n = 1 ∞ 1 ( 3 n − 1 ) 4

To determine

Whether the series is convergent or divergent.

Explanation

Given:

The series is n=1an=n=11(3n1)4.

Definition used:

The improper integral abf(x)dx is convergent if the corresponding limit exists.

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

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

Calculation:

Consider the function from given series 1(3x1)4.

The derivative of the function is obtained as follow,

f(x)=[(4)(3x1)41ddx(3x1)]       (by Chain rule)=[(4)(3x1)5(3)]=[12(3x1)5]<0

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 (1), the series is convergent if the improper integral 11(3x1)4dx is convergent.

By the definition, the improper integral is convergent if the limit exists.

Compute 11(3x1)4dx

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