Chapter 11.3, Problem 25E

### Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643

Chapter
Section

### Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
Textbook Problem

# Determine whether the series is convergent or divergent.25. ∑ n = 1 ∞ 1 n 2 + n 3

To determine

Whether the series is convergent or divergent.

Explanation

Given:

The series is n=11n2+n3 .

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 .

Calculation:

Consider the series, n=11n2+n3 . (1)

Obtain the partial fraction of 1n2+n3 .

1n2+n3=1n2(n+1)

1n2+n3=a0n2+a1n+a2(n+1) (2)

Multiply the equation by n2(n+1) and simplify the terms,

n2(n+1)n2(n+1)=a0n2(n+1)n2+a1n2(n+1)n+a2n2(n+1)(n+1)1=a0(n+1)+a1n(n+1)+a2n21=a0n+a0+a1n2+a1n+a2n21=(a1+a2)n2+(a0+a1)n+a0

Equate the coefficient of n2 , n and the constant term on both sides,

a1+a2=0a0+a1=0a0=1

Solve the above equations and obtain a0=1 ,  a1=1 and a2=1 .

Substitute 1 for a0 , -1 for a1 and 1 for a2 in equation (3),

1n2+n3=1n2+(1n)+1n+1=1n21n+1n+1

Therefore, the partial fraction is 1n2+n3=1n21n+1n+1 (3)

Substitute equation (3) in equation (1), the series is n=11n2+n3=n=1(1n21n+1n+1) .

Consider the function from the above series 1x2+x3=1x21x+1x+1

The derivative of the function is obtained as follows,

f(x)=(1)(x2)11ddx(x2)(1)(x)11ddx(x)+(1)(x+2)11ddx(x+2)=[(x2)2(2x)(x)2(1)+(x+2)2(1)]=[2xx2+(x+2)2]=[2x1x2+1(x+2)2]

Since f(x)<0 for x1 , the given function is decreasing by using the Result (2)

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