# Recursive sequence question

1. ### happyxiong531

3
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.

2. ### mathman

6,748
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: Sep 3, 2014
3. ### happyxiong531

3
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, $\frac{b_n}{1+b_n/c}\sim\frac{b_n}{1+b_n}$, then$b_n*n\rightarrow 1$.

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.

4. ### happyxiong531

3
Oh, I forget there is a condition that
the sequence should satisfy that$1-c\frac{a_1}{1+a_1}$>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

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