Convergent/Divergent Sequences

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Homework Help Overview

The problem involves determining the convergence or divergence of a sequence defined recursively, with specific initial conditions and a formula for subsequent terms. The sequence is defined as c_{1} = 4 and c_{n+1} = -\frac{c_{n}}{n^{2}} for n ≥ 1.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss evaluating specific terms of the sequence to identify a pattern or limit. There is uncertainty about how to approach the limit as n approaches infinity, particularly due to the recursive nature of the sequence.

Discussion Status

Some participants suggest examining additional terms to clarify the behavior of the sequence. There is a recognition of the alternating signs and a potential trend towards zero, although explicit consensus on convergence has not been reached.

Contextual Notes

Participants are navigating the implications of the sequence's recursive definition and the effect of the alternating signs on convergence. The original poster expresses confusion regarding the limit and the behavior of the terms.

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Homework Statement


Determine whether the sequence is convergent or divergent. Find limits for convergent sequences.

[tex]c_{1} = 4,[/tex]

[tex]c_{n+1} = -\frac{c_{n}}{n^{2}}[/tex] for [tex]n \geq 1[/tex]


Homework Equations


[tex]lim_{n\rightarrow\infty} a_{n} = L[/tex]

Where L is a number.

The Attempt at a Solution



Okay so when n=1,

[tex]c_{2} = -4[/tex]

n=2,

[tex]c_{3} = 1[/tex]

n=3,

[tex]c_{4} = -\frac{1}{9}[/tex]

I don't seem to be approaching a certain value here and I'm not sure how I can take the limit as n goes to infinity of the general term, because the general term itself depends on the previous term.

Any ideas?
 
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Why don't you try looking at a few more cases before deciding that it doesn't appear to be approaching a particular value.
 
jgens said:
Why don't you try looking at a few more cases before deciding that it doesn't appear to be approaching a particular value.

Okay.

n=4,

[tex]c_{5} = \frac{1}{144}[/tex]

n=5,

[tex]c_{5} = \frac{-1}{3600}[/tex]

Whoops! The fact that it was flipping signs confused me, this things going to zero, whether it has a negative or not!
 
Yes, it goes to zero.
 

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