Convergence of a Recursive Sequence: Proving a_n*c*n\rightarrow 1 for Positive c

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Discussion Overview

The discussion centers on the convergence of a recursive sequence defined by the relation a_{n+1}=a_n(1-c*a_n/(1+a_n)). Participants are exploring whether the product a_n*c*n approaches 1 for all positive constants c, particularly examining the case when c=1 and the implications for other values of c.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes that if a_n satisfies the recursive relation, then a_n*c*n approaches 1 for positive c, with a specific case for c=1 leading to a_n*n approaching 1.
  • Another participant questions the validity of the claim, suggesting that it implies a_n*n approaches 1/c, which seems incorrect for large c.
  • A participant argues that the proof for c=1 is valid and attempts to extend it to other values of c by redefining the sequence as b_n=ca_n, leading to a new recursive relation.
  • Concerns are raised about the strictness of the proof, with one participant noting that the sequence must satisfy a condition to ensure positivity and that c cannot be too large.
  • Plots have been made by one participant to support their claims, indicating that the conclusion holds for various values of c, including c=0.5 and c=2.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the initial claim regarding convergence. There is no consensus on the correctness of the implications for large c, and the discussion remains unresolved regarding the proof's rigor.

Contextual Notes

Participants note that there are conditions that must be satisfied for the sequence to remain positive, and the implications of varying c values introduce additional complexity that is not fully resolved.

happyxiong531
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I want to prove that
if the sequence a_n satisfy that
a_{n+1}=a_n\left(1-c\frac{a_n}{1+a_n}\right)
then a_n*c*n\rightarrow 1 for all positive c.

Like when c=1, then a_n*n\rightarrow 1,
but if c\neq 1, it's difficult to prove.
 
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What makes you believe it is true? Your question implies a_n*n ->\frac{1}{c}. Doesn't look right, especially for large c.
 
Last edited:
mathman said:
What makes you believe it is true? Your question implies a_n*n -> 1/c. Doesn't look right, especially for large c.

Thank you for you reply.
I think it's correct.
First, I can have a_n*n\rightarrow 1 when c=1, from
a_{n+1}=\frac{a_n}{1+a_n}=\frac{a_{n-1}}{1+2a_{n-1}}=\cdots=\frac{a_1}{1+(n+1)a_1}

Then, let ca_n=b_n if c\neq 1, c is some constant. we can have b_{n+1}=b_n\left(1-\frac{b_n}{1+b_n/c}\right).

Actually, it's easy to prove a_n and b_n will go to zero,
so, \frac{b_n}{1+b_n/c}\sim\frac{b_n}{1+b_n}, thenb_n*n\rightarrow 1.

I have made a plot, it's correct no matter c is larger or less than 1.
But I think my proof is not strict.
Thanks for your concern.
 
mathman said:
What makes you believe it is true? Your question implies a_n*n ->\frac{1}{c}. Doesn't look right, especially for large c.

Oh, I forget there is a condition that
the sequence should satisfy that1-c\frac{a_1}{1+a_1}>0,
so that all the elements in this sequence should be positive, and c cannot be too large.
I have made some plots like c=0.5, c=2, the conclusion is correct.
Thanks
 
Write out your complete proof.
 

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