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I have a sequence defined using the following recursion formula:

$$M_0 = N = 1000$$

$$M_{t+1} = \frac {M_t} {1 + \left( \frac1{e-1} + \frac{\sqrt{N/M_t}-1}{\sqrt{2}} \right)^{-1}}$$

I would like to know its limit when $t -> \inf$.

I note that $M_{t+1} <= M_t$, so the best candidate to the limit is 0.

However, when $M_t$ goes to 0, $M_{t+1} / M_t$ goes to 1. So, I don't know if and how I can prove that the limit is 0?

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Replace the elements of the sequence by the putative limit $M$ and calculate the candidates for it. Then try to decide between the candidates using monotony and size of the first elements. –  Phira May 9 '12 at 10:54
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up vote 3 down vote accepted

Let $f\colon[0,\infty)\to\mathbb{R}$ be defined by $$ f(x)=\frac{x} {1 + \Bigl(\frac1{e-1} +\frac{\sqrt{N/x}-1}{\sqrt{2}}\Bigr)^{-1}}. $$ Then $f$ is continuous, $f(0)=0$ and $f0<(x)<x$ for all $x>0$. The sequence $\{M_t\}$ is defined by $$ M_0=N,\quad M_{t+1}=f(M_t). $$ It follows that $$ 0<M_{t+1}<M_t\qquad \forall t\ge0. $$ Then $\lim_{t\to\infty}M_t=M\ge0$. Since $f$ is continuous, $f(M)=M$. This implies that $M\ne0$, and hence $M=0$.

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And $t^2M_t\to8N$. –  Did May 9 '12 at 13:59
    
@Didier From $f(x)=x-x^{3/2}/(10\,\sqrt{5})+O(x^2)$ and the ansatz $M_t=A/t^{-2}+o(t^{-2})$ I get $t^2M_t\to2000$. This is supported by explicit computation: $M_{100000}=2000.05096\dots$ –  Julián Aguirre May 9 '12 at 14:58
    
You are right, I made a mistake, $t^2M_t\to2N$. –  Did May 9 '12 at 15:05
    
"This implies that $M\ne0$, and hence $M=0$" - is this a typo? –  Erel Segal Halevi May 9 '12 at 20:26
    
No, it is not. We know $M\ge0$ and $f(M)=M$. But $M>0$ implies $M<f(M)$, so that $M\ne0$. If $M\ge0$ and $M\ne0$, the only possibility left is $M=0$. –  Julián Aguirre May 9 '12 at 20:48
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