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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.2012-05-09

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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) 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$\lim_{t\to\infty}M_t=M\ge0$. Since $f$ is continuous, $f(M)=M$. This implies that $M\ne0$, and hence $M=0$.

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    And $t^2M_t\to8N$.2012-05-09
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    @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$2012-05-09
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    You are right, I made a mistake, $t^2M_t\to2N$.2012-05-09
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    "This implies that $M\ne0$, and hence $M=0$" - is this a typo?2012-05-09
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    No, it is not. We know $M\ge0$ and $f(M)=M$. But $M>0$ implies $M, so that $M\ne0$. If $M\ge0$ and $M\ne0$, the only possibility left is $M=0$.2012-05-09
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    Thank you for the detailed explanation, that will help me solve similar problems in the future!2012-05-10