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Chapter 7: Problem 150

Consider the following curious expression \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\cdots}}}}\) We will use Corollary 5 to show that this actually converges to some real number. After we know it converges we can actually compute what it is. Of course to do so, we need to define things a bit more precisely. With this in mind consider the following sequence \(\left(x_{n}\right)\) defined as follows: $$ \begin{array}{c} x_{1}=\sqrt{2} \\ x_{n+1}=\sqrt{2+x_{n}}. \end{array} $$ (a) Use induction to show that \(x_{n}<2\) for \(n=1,2,3, \ldots\) (b) Use the result from part (a) to show that \(x_{n}

Short Answer

Expert verified
The sequence \(x_{n}=\sqrt{2+x_{n-1}}\) is increasing and less than 2 for all \(n\). The limit of the sequence is \(c=1+\sqrt{3}\).

Step by step solution

01

Prove that \(x_{n}

First, it is noticed that for \(x_{1}=\sqrt{2}\), the statement is true as \(\sqrt{2}<2\). Now, let's assume that for some positive integer \(k\), \(x_{k}<2\). To prove the statement is true for \(k+1\), observe the following: \[x_{k+1}=\sqrt{2+x_{k}} < \sqrt{2+2} = \sqrt{4}=2\] The inequality comes from the induction hypothesis that \(x_{k}<2\). Hence, by the principle of mathematical induction, the inequality \(x_{n}<2\) is true for all \(n\).
02

Use the result from the previous part to show that \(x_{n}

For \(x_{n}\sqrt{2+x_{n-1}}=x_{n}\] Hence, the sequence \((x_{n})\) is increasing.
03

Compute the limit of the sequence using Corollary 5

Since we have proven that \((x_{n})\) is increasing and bounded, it must be convergent by the monotone convergence theorem (Corollary 5). We denote its limit by \(c\). So, we have \(\lim_{n \to \infty}x_{n}=c\), and thus \(\lim_{n \to \infty}x_{n+1}=c\) as well. Since \(x_{n+1}=\sqrt{2+x_{n}}\), we can create the equation \(c=\sqrt{2+c}\). Squaring both sides gives \(c^2=2+c\), simplifying yields \(c^2-c-2=0\). The roots of this quadratic equation are \(c=1-\sqrt{3}\) and \(c=1+\sqrt{3}\). But, since the sequence \((x_{n})\) is increasing and \(x_{1}=\sqrt{2}<2\), the limit \(c\) must be the larger root \(c=1+\sqrt{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convergence of Sequences
When we discuss the convergence of sequences in real analysis, we're looking at whether the terms of a sequence approach a specific value as they progress to infinity. For the sequence given in the problem, defined by \(x_1 = \sqrt{2}\) and \(x_{n+1} = \sqrt{2+x_n}\), we want to demonstrate that these terms settle down to a single number.

A sequence \((x_n)\) converges to a limit \(L\) if, for every \(\epsilon > 0\), there is a natural number \(N\) such that for all \(n > N\), \(|x_n - L| < \epsilon\). In simple terms, the terms get arbitrarily close to \(L\) as \(n\) becomes very large.
  • Our example sequence is being crafted in such a way that each \(x_{n+1}\) is built upon \(x_n\), caused by the repeating square root structure.
  • By showing it's both increasing and bounded, and using the monotone convergence theorem, we confirm it converges.
Mathematical Induction
Mathematical induction is a powerful tool to prove statements about integers, often involving sequences. It works like a domino effect, showing that if one piece falls (a statement holds for one number), the rest will follow.

To prove a property using induction, we follow two steps:
1. **Base Case**: Prove the statement is true for the first number, typically \(n=1\).2. **Inductive Step**: Assume the statement holds for an arbitrary number \(k\), and then show it must hold for \(k+1\).
  • For our sequence, the base case was \(x_1 = \sqrt{2} < 2\), which clearly holds.
  • In the inductive step, under the assumption \(x_k < 2\), we showed \(x_{k+1} = \sqrt{2+x_k} < 2\) as well, concluding the proof.
Monotone Convergence Theorem
The Monotone Convergence Theorem is a cornerstone in real analysis, especially when dealing with sequences.

It states that if a sequence is both monotonic (consistently increasing or decreasing) and bounded, it must converge to a limit.

For our sequence:
  • We proved it's increasing using the fact that \(x_n < x_{n+1}\).
  • We also showed it's bounded above by 2 using induction.
With the monotone convergence theorem, these two facts together guarantee that the sequence converges. We found the limit by setting \(x_n\) and \(x_{n+1}\) to \(c\), solving the equation \(c = \sqrt{2+c}\) to get \(c = 1 + \sqrt{3}\), given the sequence bounds.

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Most popular questions from this chapter

The purpose of this problem is to show that $$ \lim _{n \rightarrow \infty}\left(\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\ln (n+1)\right) $$ exists. (a) Let \(x_{n}=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\ln (n+1) .\) Use the following diagram to show \(x_{1} \leq x_{2} \leq x_{3} \leq \cdots\) (b) Let \(z_{n}=\ln (n+1)-\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+1}\right) .\) Use a similar diagram to show that \(z_{1} \leq z_{2} \leq z_{3} \leq \cdots\) (c) Let \(y_{n}=1-z_{n} .\) Show that \(\left(x_{n}\right)\) and \(\left(y_{n}\right)\) satisfy the hypotheses of the nested interval property and use the NIP to conclude that there is a real number \(\gamma\) such that \(x_{n} \leq \gamma \leq y_{n}\) for all \(n .\) (d) Conclude that \(\lim _{n \rightarrow \infty}\left(\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\ln (n+1)\right)=\gamma\).

Let \(S \subseteq \mathbb{R}\) and let \(T=\\{-x \mid x \in S\\}\). (a) Prove that \(b\) is an upper bound of \(S\) if and only if \(-b\) is a lower bound of \(T\). (b) Prove that \(b=\sup S\) if and only if \(-b=\inf T\).

Use the Least Upper Bound Property to prove the Nested Interval Property. That is, assume that every non-empty subset of the real numbers which is bounded above has a least upper bound; and suppose that we have two sequences of real numbers \(\left(x_{n}\right)\) and \(\left(y_{n}\right),\) satisfying: 1\. \(x_{1} \leq x_{2} \leq x_{3} \leq \ldots\) 2\. \(y_{1} \geq y_{2} \geq y_{3} \geq \ldots\) 3\. \(\forall n, x_{n} \leq y_{n}\)

Does \(\mathbb{Q}\) satisfy the Archimedean Property and what does this have to do with the question of taking the Archimedean Property as an axiom of completeness?

Suppose \(\lim _{n \rightarrow \infty} x_{n}=c .\) Prove that \(\lim _{k \rightarrow \infty} x_{n_{k}}=c\) for any subsequence \(\left(x_{n_{k}}\right)\) of \(\left(x_{n}\right) .\) [Hint: \(\left.n_{k} \geq k .\right]\) A very important theorem about subsequences was introduced by Bernhard Bolzano and, later, independently proven by Karl Weierstrass. Basically, this theorem says that any bounded sequence of real numbers has a convergent subsequence.
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