   Chapter 11.6, Problem 51E

Chapter
Section
Textbook Problem

# Given any series Σ an we define a series ∑ a n + whose terms are all the positive terms of Σ an and a series ∑ a n − whose terms are all the negative terms of Σ an. To be specific, we let a n + = a n + | a n | 2   a n − = a n − | a n | 2 Notice that if an > 0, then a n + = a n and a n − = 0 whereas if an < 0, then a n − = a n and a n + = 0 . (a) If Σ an is absolutely convergent, show that both of the series Σ   a n + and Σ   a n − are convergent. (b) If Σ an is conditionally convergent, show that both of the series Σ   a n + and Σ   a n − are divergent.

(a)

To determine

To show: The series an+ and an are convergent.

Explanation

Given:

The series an  is absolutely convergent.

Definition used:

A series an  is absolutely convergent if |an| is convergent.

Theorem used:

“If a series an is absolutely convergent, then it is convergent.”

Result used:

(1) If the series an and bn are convergent series then an+bn=an+bn.

(2) Sum of two convergent series is convergent.

(3) Difference of two convergent series is convergent.

Calculation:

Consider the series an+ whose terms are positive terms and the series an whose terms are negative terms of an.

The series an+ can be expressed as follows,

an+ =an+|an|2=12[(an+|an|)]=12[an+|an|]

Since an is absolutely convergent, |an| is convergent is convergent by the definition

(b)

To determine

To show: The series an+ and an are divergent

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