   Chapter 11.1, Problem 82E

Chapter
Section
Textbook Problem

# Show that the sequence defined by a 1 = 2   a n + 1 = 1 3 − a n satisfies 0 < an ≤ 2 and is decreasing. Deduce that the sequence is convergent and find its limit.

To determine

To show: The sequence {an} is increasing and an<3 for all n . Also, deduce that the sequence is convergent and then obtain the limit of the sequence.

Explanation

Given:

The sequence is a1=2, an+1=13an (1)

Theorem used: Monotonic Sequence Theorem

“If the sequence is bounded and monotonic, then the sequence is convergent.”

Proof: Induction Method

Claim: Let Pn be the statement that an+1an and 0<an2 . (2)

Base case: n=1

Substitute 1 for n in equation (2),

a1+1a1 and 0<a12a2a1 and 0<a12

Obtain the value of a2 .

Substitute 1 for n in equation (1),

a1+1=13a1a2=132 [a1=2]=11=1

Thus, the value of a2=1 .

Since the value of a1=2 and a2=1 , a2a1  and  0<a12 .

Therefore, the claim is true for n=1 .

Induction hypothesis: n=k

Assume that the claim is true when n=k .

If Pk is the statement, then ak+1ak and 0<ak2 is true.

Inductive step: n=k+1

To prove that the claim is true when n=k+1

That is, if Pk+1 is the statement, then a(k+1)+1ak+1 and 0<ak+12 is true.

Substitute k+1 for n in equation (1),

a(k+1)+1=13ak+1 (3)

From induction hypothesis,

ak+1akak+1>ak

Add by 3 on both sides of the inequality.

3ak+1>3ak

13ak+113ak (4)

Substitute the inequality (4) in the equation (3) and then plug in 13ak for ak+1 .

a(k+1)+113ak

a(k+1)+1ak+1 (5)

Consider ak+1 , then apply equation (1) and the induction hypothesis 0<ak2

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