Converging Sequence Homework: Determine Limit & Explain

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Homework Help Overview

The discussion revolves around determining the convergence of a sequence defined by the expression a_n = √(n² + 1) - n. Participants are tasked with finding the limit if it converges and providing detailed explanations for their claims.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the Squeeze Theorem and L'Hopital's Rule to analyze the sequence. There is a debate about the effectiveness of these methods, particularly regarding the application of the Squeeze Theorem and the justification for choosing bounds. Some participants question the logical basis for selecting certain limits in the context of the Squeeze Theorem.

Discussion Status

The discussion is ongoing, with participants offering various approaches and questioning each other's reasoning. Some guidance has been provided, such as the suggestion to rationalize the expression, but no consensus has been reached on the best method to prove convergence.

Contextual Notes

Participants express concerns about justifying their choices for the boundaries in the Squeeze Theorem and the need for logical reasoning behind their assumptions. There is also mention of imposed homework rules requiring detailed explanations for claims made.

Lucretius
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Homework Statement



Determine whether each of the sequences converges or diverges. If it converges, find its limit. Explain with sufficient details each claim.

Homework Equations


(k) [tex]a_n=\sqrt{n^2+1}-n[/tex]

The Squeeze Theorem (If [tex]b_n<a_n<c_n[/tex], and both [tex]\displaystyle\lim_{x\rightarrow\infty}a_n, c_n = L[/tex], then [tex]\displaystyle\lim_{x\rightarrow\infty}b_n=L[/tex])

The Attempt at a Solution



I had the choice of deciding between the squeeze theorem and L'Hopital. L'Hopital seemed useless because I don't see a way to get this into the form of a quotient. I know the sequence is convergent to 0. Figured it would be best to try the Squeeze theorem, as it's really my only other option.

I know that [tex]0 \leq \sqrt{n^2+1}-n[/tex], but after struggling for a while, I still have no idea how to get a larger limit on the other side equal zero (without arbitrarily choosing one, 1/x for instance works, but there's no logical way to get from my function to 1/x...) I know that [tex]n+1>\sqrt{n^2+1}[/tex], but the result is n+1-n=1, and the limit of 1 is 1, period. Likewise, I can get rid of the 1, but I'll have n left over, and the limit of n is infinity! Got any good tips on getting the other side of the squeeze theorem inequality?
 
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Lucretius said:

Homework Statement



Determine whether each of the sequences converges or diverges. If it converges, find its limit. Explain with sufficient details each claim.

Homework Equations


(k) [tex]a_n=\sqrt{n^2+1}-n[/tex]

Multiply by [tex]\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}[/tex] and watch the magic.
 
You stated the Squeeze theorem incorrectly, but I suppose you know the correct statement. I don't know what you mean by there being "no logical way" to get 1/n, but if 1/n works, then it works, so what's the problem? 1/n converges to 0 as n goes to infinity, if you can prove that [itex]\sqrt{n^2+1}-n\leq n^{-1}[/itex] for all n, then you're done. There's an easier way in my opinion, but do it your way first, and once you have your own answer I can give you mine.
 
Ah thanks quasar; I should have thought of rationalization. I need to remember that everytime I see a square root...

When I meant "logical way" AKG, I meant that, just by looking at the equation, we can't conclude that 1/x is bigger always. I had to look at a calculator. I have to justify my choice for the boundaries of the squeeze theorem. However, quasar pointed out I could just rationalize the function and end up finding out that 1/n is bigger.
 
Lucretius said:
When I meant "logical way" AKG, I meant that, just by looking at the equation, we can't conclude that 1/x is bigger always.
No, but you should be able to prove it.
 

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