Chapter 9, Problem 2PS

### Calculus: Early Transcendental Fun...

6th Edition
Ron Larson + 1 other
ISBN: 9781285774770

Chapter
Section

### Calculus: Early Transcendental Fun...

6th Edition
Ron Larson + 1 other
ISBN: 9781285774770
Textbook Problem

# Using Sequences(a) Given that lim n → ∞ a 2 n and lim n → ∞ a 2 n + 1 = L , show that{ a n } is convergent and lim n → ∞ a n = L .(b) Let a 1 = 1 and a n + 1 = 1 + 1 1 + a n Write out the Firsteight terms of { a n }. Use part (a) to show that lim n → ∞ a n = 2 This gives the continued fraction expansion 2 = 1 + 1 2 + 1 2 + ⋯ .

(a)

To determine

To prove: The sequence {an} is convergent and limxan=L.

Explanation

Given:

The limit values of sequence are, limxa2n=L and limxa2n+1=L.

Formula used:

The limit can be written as: limxaf(x)=f(a).

Proof:

Consider the sequence bn=an.

If M is a bound below for an, that is M<an for all n,

Then, bn=an<M,

So, bn is bounded above.

Also, it can be seen that anan+1 that anan+1 that is bnbn+1.

Therefore, the {bn} converges to some limit L by the monotone convergence theorem.

Thus, the limit law sequence {an} is also converges to L.

Let, ε>0 be given. limxa2n=L means there exists M1 such that |a2nL|<ε for n>M1

(b)

To determine

To calculate: The first eight terms of sequence {an}.

### 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