Chapter 11.6, Problem 52E

### Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

Chapter
Section

### Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

# Prove that if ∑ a n is a conditionally convergent series and r is any real number, then there is a rearrangement of ∑ a n whose sum is r. [Hints: Use the notation of Exercise 51. Take just enough positive terms a n + so that their sum is greater than r. Then add just enough negative terms a n − so that the cumulative sum is less than r. Continue in this manner and use Theorem 11.2.6.]

To determine

To prove:

If an is conditionally convergent series and r is any real number, then there is rearrangement of an whose sum is r.

Explanation

1) Concept:

The partial sum:

The given series n=1an=a1+a2+a3+, let sn denote its nth partial sum,

sn=i=1nai=a1+a2++an

If sequence {sn} is convergent and limnsn=s exists a real number, then the series an is called convergent.

2) Theorem:

If the series n=1an is convergent, then limnan=0

3) Calculation:

Let, bn be the rearranged series.

an+ represents positive series and an- represents negative series, where an+ and an- can be defined as follows,

For an>0, an+=an and for an<0, an+=0

For an>0, an-=0 and for an<0, an-=an

Recall from previous exercise that an+ is a divergent series. Moreover each term of it is positive. So there exists some k1 such that, the partial sum Sk1=i=1k1ai+>r.  And i=1k1-1ai+r.

Now since an- are all negative and forms a divergent series there exists some l1 such that the partial sum Sk1+l1=i=1k1ai++i=1l1ai-<r. And Sk1+l1-1 r

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started