   Chapter 11.2, Problem 32E

Chapter
Section
Textbook Problem

# Determine whether the series is convergent or divergent. If it is convergent, find its sum.32. ∑ n = 1 ∞ [ ( − 0.2 ) n + ( 0.6 ) n − 1 ]

To determine

Whether the series is convergent or divergent and obtain the sum if the series is convergent.

Explanation

Given:

The series is n=1[(0.2)n+(0.6)n1] .

Result used:

The geometric series n=1arn1 (or) a+ar+ar2+ is convergent if |r|<1 and its sum is a1r , where a is the first term and r is the common ratio of the series.

Calculation:

The given series can be expressed as follows,

n=1[(0.2)n+(0.6)n1]=n=1(0.2)n+n=1(0.6)n1 (1)

Here, both the series n=1(0.2)n and n=1(0.6)n1 are geometric series.

Obtain the sum of the geometric series n=1(0.2)n if the series is convergent.

Consider the series, n=1(0.2)n=(0.2)+(0.2)2+(0.2)3+ .

Here, the first term of the series is, a1=0.2 and the common ratio of the series is,

r1=(0.2)2(0.2)=0.2

The absolute value of r1 is,

|r1|=|0.2|=0.2<1

Since |r1|<1 and by the Result stated above, the series is convergent.

Obtain the sum of the series.

Compute the value of n=1(0.2)n=a11r1 .

Since a1=0.2 and r1=0.2 ,

a11r1=0.21(0.2)=0.21+0.2=0.21.2=16

Thus, the sum of the series is n=1(0.2)n=16

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