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Let be the following sequence $x_{n}$, $n\geq0$, $0<x_{0}<1$ that satisfies the following recurrence: $$ x_{n+1}=x_{n}-x_{n}^2+x_{n}^3-x_{n}^4+x_{n}^5-x_{n}^6$$

Having given this, i'm required to calculate: $$\lim_{n\rightarrow\infty} n x_{n} $$

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1 Answer 1

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The recurrence yields $$ \frac{x_{n+1} }{x_n} = (1-x_n) (1+x_n^2 + x_n^4) .$$

For $z\in (0,1)$ we have $ 0< (1-z)(1+z^2+z^4) < 1$ so $x_n$ is montonically decreasing and non-negative. Also note that since $x_n$ is a non-negative sequence we have $0\leq x_n \leq 1$ for all $n\in \mathbb{N}$ (for if $x_k>1$ then $x_{k+1}<0$). Thus $x_n\to x$ where $x\in[0,1].$ It must satisfy $$x=x(1-x)(1+x^2+x^4)$$ which can be written $$ x\left((1-x)(1+x^2+x^4) -1\right)=0.$$ Thus $\displaystyle \lim_{n\to \infty} x_n=0$ and consequentially, $\displaystyle \lim_{n\to\infty} \frac{x_{n+1}}{x_n} = 1.$

Now we calculate $$ \frac{(n+1) - n}{\frac{1}{x_{n+1}} - \frac{1}{x_n} }= \frac{x_{n+1} x_n}{x_n - x_{n+1}}= \frac{x_{n+1}x_n}{x_n^2-x_n^3 +x_n^4 -x_n^5 +x_n^6} $$

$$= \frac{x_{n+1}/x_n }{1-x_n +x_n^2 - x_n^3 + x_n^4}\to 1.$$

Hence, by the Stolz–Cesàro theorem we conclude $\displaystyle \lim_{n\to\infty} nx_n =1.$

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great solution! Thanks. –  Chris's sis Jun 2 '12 at 14:40

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