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Chapter 7: Q. 37 (page 657)

Conditional and absolute convergence: For each of the series that follow, determine whether the series converges absolutely, converges conditionally, or diverges. Explain the criteria you are using and why your conclusion is valid.

鈭�k=0鈭�(-1)k3k

Short Answer

Expert verified

The series 鈭�k=0鈭�(-1)k3kconverges conditionally.

Step by step solution

01

Step 1. Given Information.

The series:

鈭�k=0鈭�(-1)k3k

02

Step 2. By Alternating Series Test.

According to the Alternating Series Test, the sequence ak+1<akfor every . Then the alternating series ak+1,akboth converges.

03

Step 3. Find ak+1.

ak=13kak+1=13k+1

So the sequence is monotonic decreasing sequence.

04

Step 4. Find limk→∞ak.

limk鈫�鈭�ak=limk鈫�鈭�(13k)=0

So the series converges.

05

Step 5. Find if it converges conditionally or absolutely.

According to the convergence or divergence of the geometric series, the series ak=鈭�k=0鈭�13kwhere r<1, the serie converges to 11-r.

11-r=11-13=123=32

Hence the series converges conditionally.

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Most popular questions from this chapter

Explain why a function a(x) has to be continuous in order for us to use the integral test to analyze a series 鈭�k=1鈭�akfor convergence.

Express each of the repeating decimals in Exercises 71鈥�78 as a geometric series and as the quotient of two integers reduced to lowest terms.

0.99999...

Consider the series

Fill in the blanks and select the correct word:

Iflimk鈫�鈭�akbk=鈭�and鈭�k=1鈭�_____divergesthen鈭�k=1鈭�_____(converges/diverges).

Use the divergence test to analyze the given series. Each answer should either be the series diverges or the divergence test fails, along with the reason for your answer.

鈭�k=1鈭�1k

For a convergent series satisfying the conditions of the integral test, why is every remainder Rnpositive? How can Rnbe used along with the term Sn from the sequence of partial sums to understand the quality of the approximation Sn?
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