(a) Show that if limn →a a2 n=L and limn →∞ a2 n+1=L, then {an} is convergent and limn →∞ an=L .

(a) Show that if limn →a a2 n=L and limn →∞ a2 n+1=L, then...

(a) Show that if $\lim _{n \rightarrow a} a_{2 n}=L$ and $\lim _{n \rightarrow \infty} a_{2 n+1}=L,$ then $\left\{a_{n}\right\}$ is convergent and $\lim _{n \rightarrow \infty} a_{n}=L .$ (b) If $a_{1}=1$ and $a_{n+1}=1+\frac{1}{1+a_{n}}$ find the first eight terms of the sequence $\left\{a_{n}\right\} .$ Then use part (a) to show that $\lim _{n \rightarrow \infty} a_{n}=\sqrt{2} .$ This gives the continued fraction expansion $\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\cdots}}$

(a) Show that if $\lim _{n \rightarrow a} a_{2 n}=L$ and $\lim _{n \rightarrow \infty} a_{2 n+1}=L,$ then $\left\{a_{n}\right\}$ is convergent and $\lim _{n \rightarrow \infty} a_{n}=L .$
(b) If $a_{1}=1$ and $a_{n+1}=1+\frac{1}{1+a_{n}}$
find the first eight terms of the sequence $\left\{a_{n}\right\} .$ Then use part (a) to show that $\lim _{n \rightarrow \infty} a_{n}=\sqrt{2} .$ This gives the continued fraction expansion $\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\cdots}}$

Key Concepts

Sequence Convergence

This concept deals with the idea that a sequence approaches a single number (limit) as its index goes to infinity. In a convergent sequence, for every tolerance, there exists a point in the sequence such that all subsequent terms remain within that tolerance of the limit. Understanding this definition is fundamental to analyzing the behavior of sequences in analysis.

Subsequence Convergence

Subsequences are derived by taking certain elements from the original sequence, such as every other term. A key theorem states that if every subsequence converges to the same limit, then the entire sequence must also converge to that limit. This is particularly useful when the sequence can be naturally split into parts (such as even and odd indexed terms) that are easier to analyze individually.

Recursively Defined Sequences

A recursively defined sequence is one where each term is defined as a function of preceding terms. Analyzing such sequences involves computing initial terms to look for patterns and studying the recurrence relation to determine long-term behavior. Many recursive sequences converge to a limit, and understanding their structure is essential in many areas of mathematics.

Fixed Point Analysis

In the study of recursively defined sequences, a fixed point is a value that remains unchanged under the recurrence relation. If a sequence converges, its limit must satisfy the fixed point equation derived from the recurrence. Finding and analyzing such fixed points often provides a direct method to calculate the limit of the sequence.

Continued Fraction Expansion

Continued fraction expansion represents numbers through a nested sequence of fractions. It is particularly useful for expressing irrational numbers and understanding their properties. Analyzing the pattern in a continued fraction can offer insights into the convergence of the generated sequence and connects recursive definitions to classical representations of numbers.

Transcript

00:01 In this problem, we're given that the sequence a2n and also the sequence a2n plus 1, that these both converge to the same limit l.

00:18 So by definition, let's go ahead and let epsilon be a positive number.

00:27 So by this fact here, there exists some positive number, let's call it n1, such that if we go ahead and take n larger than this, this.

01:01 And similarly, there's another integer, we can say n2, such that when we take n larger than this n2, we get a similar inequality, but this time involving 2n plus 1.

01:31 Now, the next step is to choose little n large enough so that we can have both of these at the same time.

01:41 So if we have both of these inequalities at the same time, if you notice here, this is all the even ends, and then this is all the odd values.

01:53 And so every number end that we choose is going to be even or odd.

01:59 And we'd like to say eventually, this is the inequality that we desire.

02:07 So we'll go ahead and let n, we'll take n to be really large.

02:11 Let's take it to be larger than two times, let me take the max, the larger of n1 and n2.

02:22 And let me add one to that.

02:25 The reason i'm putting the two times and then the plus one is because of i want to make sure that if i multiply this n here by 2 and add 1, that i could use both of these inequalities over here.

02:46 If you notice these have 2n and 2n plus 1, whereas the one down here only has an n.

02:52 That's why i have to make up for it by including the two and the plus one just to ensure.

02:59 So i'll just put here in parentheses, ensures we can use.

03:11 And as we mentioned now, because now that we fix little n to be larger than this number over here, further, every value of n will be even or odd...

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