State why lim Xn exists, and find the limit

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Homework Help Overview

The problem involves a sequence defined by x1 = √2 and xn+1 = √(2 + xn). Participants are tasked with demonstrating properties of the sequence, including showing that it is increasing and bounded, and ultimately discussing the existence of the limit and finding its value.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship that the limit must satisfy and how to substitute the limit into the recursive definition of the sequence. There are attempts to derive the limit through algebraic manipulation of the equation.

Discussion Status

The discussion includes various attempts to find the limit, with some participants expressing uncertainty about the correctness of their findings. There is an ongoing exploration of the implications of the limit being 2 and questioning the validity of this conclusion.

Contextual Notes

Participants are navigating through the requirements of mathematical induction and the implications of the sequence being bounded and increasing. There is a focus on understanding the conditions under which the limit exists.

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Homework Statement



Let x1 = √2, xn+1 = √(2+xn). Use mathematical induction to show that xn < xn+1. Next show that if xn ≥ 2, then xn-1 ≥ 2 also. How do you conclude from this that xn ≤ 2 for all n? State why lim xn exists, and find the limit.

Homework Equations



Upper bounds, limits, etc.

The Attempt at a Solution



I've done everything up to the point of "find the limit." Help?
 
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You have

\begin{align*}<br /> x_1 &amp; = \sqrt{2} \\<br /> x_2 &amp; = \sqrt{2+\sqrt{2}} \\<br /> x_3 &amp; = \sqrt{2+\sqrt{2+\sqrt{2}}} \\<br /> x_4 &amp; = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} \\<br /> &amp; \vdots<br /> \end{align*}

Can you see what relationship the limit must satisfy?
 
Hi Jamin2112! :smile:

If the limit exists, and is x, then you can substitute x for xn and xn+1 in the same equation. :wink:
 
tiny-tim said:
Hi Jamin2112! :smile:

If the limit exists, and is x, then you can substitute x for xn and xn+1 in the same equation. :wink:

So, since xn+1 = √(2 + xn) and lim xn+1 = lim xn, we can denote the limit A, and solve for A as such:

A = √(2+A)
A2 = 2 + A
A2 - A - 2 = 0
(A-2)(A+1) = 0
A = 2 , -1

Therefore A = 2.

That seems wrong.
 
2 = √(2 + 2) … what's wrong with that? :smile:
 
\sqrt{2}= 1.414...
\sqrt{2+ \sqrt{2}}= \sqrt{3.414..}= 1.84776
\sqrt{2+ \sqrt{2+ \sqrt{2}}= \sqrt{3.84776}= 1.96157
\sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2}}}= \sqrt{3.96157}= 1.99037

Why would "2" not seem right?
 

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