   Chapter 11.3, Problem 18E

Chapter
Section
Textbook Problem

# Determine whether the series is convergent or divergent.18. ∑ n = 1 ∞ 1 n 2 + 2 n + 2

To determine

Whether the series is convergent or divergent.

Explanation

Given:

The series is n=11n2+2n+2 .

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 .

Definition used:

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

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

Calculation:

Consider the function from given series 1x2+2x+2 .

The derivative of the function is obtained as follows,

f(x)=(1)(x2+2x+2)11ddx(x2+2x+2)   (Chain rule)=((x2+2x+2)2(2x+2))=2x+2(x2+2x+2)2=2(x+1)(x2+2x+2)2

Since f(x)<0   for x1 , 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 11x2+2x+2dx is convergent.

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

Compute 11x2+2x+2dx as shown below

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