Limit of a Sequence with (-1)n and Convergence Analysis

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Homework Help Overview

The discussion revolves around the limit of a sequence involving the term (-1)n and its convergence properties. Participants are exploring the nature of the sequence defined by this term and its behavior as n approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are considering the sequence defined as xn = n[1+(-1)n] + (1/n) and questioning its convergence and accumulation points. There is a focus on writing out the first few terms to analyze the sequence's behavior.

Discussion Status

Some participants have noted the potential for the sequence to have one accumulation point, while questioning whether it converges. A proof regarding the convergence of sequences of the form (-1)^n a_n is also presented, suggesting a deeper exploration of the conditions under which convergence may occur.

Contextual Notes

There is an ongoing examination of the implications of the sequence's structure, particularly regarding the conditions for convergence and the nature of its accumulation points. Participants are also considering the constraints imposed by the sequence's alternating nature.

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Homework Statement



... I have (-1)n in a sequence and I'm trying to find the limit of that sequence?

Homework Equations



Obviously we have -1, 1, -1, 1, ...

The Attempt at a Solution



You see, I'm trying to consider the case xn = n[1+(-1)n] + (1/n), to consider whether the sequence has just one accumulation point and is convergent.
 
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The sequence does seem to have one accumulation point, but could it converge? Write out the first few terms.
 
Tedjn said:
The sequence does seem to have one accumulation point, but could it converge? Write out the first few terms.

xn = {1, 9/4, 1/3, 65/8, 1/5, 125/12, 1/7, ...}
 
If a sequence is of the form [itex](-1)^n a_n[/itex] where [itex]a_n\ge 0[/itex] for all n, then either the sequence does not converge or it converges to 0.

Proof: suppose it converged to a> 0. Let [itex]\epsilon= a/2[/itex]. Then for any N> 0, there exist odd n> N so that [itex](-1)^na_n< 0[/itex] from which [itex]|a- (-1)^n a_n|> a[/itex] and not less than [itex]\epsilon= a/2[/itex].

You do the case for a< 0.
 

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