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Suppose we have $z_{n+1}=\frac{z_{n}^2}{1+cz_{n}}$ where $c>1$ and $z_{1}>0$. What can we say about $z_{n}$? Can we find an explicit formula? Can we at least get an approximation of the form $c_{1}a_{1}(n)+c_{2}a_{2}(n)+\mathcal O(a_{3}(n))$ for some constants $c_i$ and some sequences $a_{i}$ with decreasing order of magnitude?

Edited: the index of the sequence in $\mathcal O$.

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Note that $\frac{z_{n}}{z_{n+1}}=\frac{1+c z_n}{z_n}=c+\frac1{z_n}>c>1$, hence the sequence is strictly decreasing. Ultimately, $z_n\to 0$ with $z_{n+1}\approx z_n^2$. Letting $a=\lim_{n\to\infty}\frac{\ln z_n}{2^n}$ (if it exists?) it seems reasonable to compare $z_n$ with $\exp(2^na)$. By the way, note that $c_1a_1(n)+c_2a_2(n)+{\mathcal O}(a_1(n))=c_2a_2(n)+{\mathcal O}(a_1(n))$. –  Hagen von Eitzen Sep 15 '12 at 12:04

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Introduce $\gamma=\gamma(c,z_1)$ with $0\lt\gamma\lt1$, as $$ \gamma=z_1^{1/2}\cdot\prod_{n\geqslant1}(1+cz_n)^{-1/2^n}. $$ Then $z_n^{1/2^n}\to\gamma$ decreasingly, and, considering $\delta\lt\gamma$ defined by $\delta=\gamma^3/4$, $$ z_n=\gamma^{2^n}+c\cdot\delta^{2^n}+o(\delta^{2^n}). $$ To prove this, one can start from the formula $$ z_n^{1/2^n}=\gamma\cdot\prod_{k\geqslant n+1}(1+cz_k)^{1/2^k}. $$

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