   Chapter 11.6, Problem 49E

Chapter
Section
Textbook Problem

Prove the Root Test. [Hint for part (i): Take any number r such that L < r < 1 and use the fact that there is an integer N such that | a n | n < r whenever n ≥ N.]

To determine

To prove: The Root Test.

Explanation

Given:

limn|an|n=L .

Result used:

(1)“Suppose that an and bn are the series with positive terms,

(a) If bn is convergent and anbn for all n , then an is also convergent.

(b) If bn is divergent and anbn for all n , then an is also divergent.”

(2) The geometric series n=1arn1 is convergent if |r|<1 and divergent if |r|>1 .

Definition used:

“A series an is called absolutely convergent if the series of absolute values |an| is convergent.”

Calculation:

Consider the given series n=1|an| .

Case (i): L<1

By given statement as follows,

limn|an|n=L<1

That is, limn|an|n<1 .

Choose the number r such that L<r<1 .

That is, there exists a number N such that |an|n<r for all nN .

|an|n<r|an|<rn<1

n=1|an|<n=1rn

Thus, by the Result (2) stated above, the series n=1rn converges with 0<|r|<1 .

Use the Result (1), n=1|an| is convergent.

By the definition, n=1an is absolutely convergent if L<1 .

Case (ii): L>1

By given statement as follows,

limn|an|n=L>1

That is, choose any number N such that limn|an|n>1 for all nN .

|an|>1 for all nN

That is, limnan0 .

The series n=1an diverges by Test for Divergence.

Therefore, the series n=1an diverges if L>1 .

Case (iii): L=1

Consider the convergent series n=1an=n=11n2 and divergent series n=1bn=n=11n .

limn|an|n=limn|(1n2)|1n=limneln(1n2)1n=limne1nln(1n2)

limn|an|n=elimn1nln(1n2) (1)

Obtain limn1nln(1n2)

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