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Let $0< x_1 < \frac{1}{q}$, where $0 < q \le 1$, and $x_{n+1}=x_{n}(1 - qx_n)$ for $n\in \mathbb{N}$.
Prove that $\lim\limits_{n \rightarrow \infty}{nx_n = \frac{1}{q}}$.

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    You are correct, then you simplify $\frac{1}{q}$ from both sides and you get an equivalent problem, except now without the parameter $q$. and you have 0< y_1 <1,\, y_{n+1}=y_n(1-y_n) and you need to show $ \lim_ { n \rightarrow \infty} n y_n =1$2012-10-05

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Let $f(x) = x (1 - q x)$. For $0 < x < 1/q$ we have $0 < 1 - q x < 1$ so $0 < f(x) < x$. Since the only fixed point of the continuous function $f$ is $0$, the limit of the decreasing sequence $x_n$ is $0$. Now if $a_n = 1/x_n$, we have $a_{n+1} - a_n = \dfrac{q}{1 - q x_n}$. Take $b_n = n$ in the statement of the Stolz–Cesàro theorem as in http://en.wikipedia.org/wiki/Stolz_theorem