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Chapter 3: Problem 4

Let \(x_{1}:=1\) and \(x_{n+1}:=\sqrt{2+x_{n}}\) for \(n \in \mathbb{N}\). Show that \(\left(x_{n}\right)\) converges and find the limit.

Short Answer

Expert verified
The sequence \(x_{n}\) converges to 2 as \(n\) tends to infinity.

Step by step solution

01

Base Case Verification

Start by verifying the constraints for the base case of mathematical induction i.e., n = 1. Here \(x_{1}\) is defined as 1, which is clearly true.
02

Inductive Step

Now, proceed to the inductive step. Let's assume for some arbitrary natural number \(k\), the statement holds true i.e., \(x_{k+1} = \sqrt{2+ x_{k}}\). Our goal is to prove the statement holds for \(x_{k+2}\) using the assumption.
03

Apply Inductive Hypothesis

Replace \(k+1\) by 'k' in \(x_{k+2}\) equation. Based on our assumption, we can replace \(x_{k+1}\) by \(\sqrt{2+x_{k}}\) in \(x_{k+2}\) equation. So, \(x_{k+2} = \sqrt{2+\sqrt{2+ x_{k}}}\). This demonstrates the sequence.
04

Convergence Proving

To show the convergence, we have to show that the sequence is monotonically increasing and bounded above. Monotonic property is easy to show here as we always add 2 and take a square root which will never decrease the value of \(x_{n}\). For the upper bound, we can see by induction that \(x_{n} < 2\) for all \(n\), and thus the sequence is bounded above by 2.
05

Finding the Limit

To find the limit of the converging sequence \(x_n\), we let it be 'L'. We know that \(x_{n+1}\) tends to L as \(n\) tends to infinity. We therefore substitute \(x_{n+1}\) and \(x_n\) by 'L' in the equation \(x_{n+1} = \sqrt{2 + x_{n}}\). Solving the resultant equation gives the limit of the sequence \(x_n\).

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Most popular questions from this chapter

(a) Give an example of a convergent sequence \(\left(x_{n}\right)\) of positive numbers with \(\lim \left(x_{n}^{1 / n}\right)=1\). (b) Give an example of a divergent sequence \(\left(x_{n}\right)\) of positive numbers with \(\lim \left(x_{n}^{1 / n}\right)=1\). (Thus, this property cannot be used as a test for convergence.)

Show that if \(\left(x_{n}\right)\) is an unbounded sequence, then there exists a properly divergent subsequence.

Show that if \(x_{n} \geq 0\) for all \(n \in \mathbb{N}\) and \(\lim \left(x_{n}\right)=0\), then \(\lim \left(\sqrt{x_{n}}\right)=0\).

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, and if \(b_{n}:=\left(a_{1}+\cdots+a_{n}\right) / n\) for \(n \in \mathbb{N}\), then show that \(\sum b_{n}\) is always divergent.

Use the Cauchy Condensation Test to discuss the \(p\) -series \(\sum_{n=1}^{\infty}\left(1 / n^{p}\right)\) for \(p>0\).
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