[再寄小读者之数学篇](2014-05-25 非线性递归数列的敛散性)

时间：2014-05-25 11:24:34 阅读：364 评论：0 收藏：0 [点我收藏+]

数列 $\begin{Bmatrix} {x}_{n} \end{Bmatrix}$ 满足如下定义:

a > 0, b > 0; x 1 = a, x 2 = b; x n + 2 = 2 + 1 x 2 n + 1 + 1 x 2 n, n \geq 1.

$a>0,\quad b>0; \qquad {x}_{1}=a,\quad{x}_{2}=b ;\qquad {x}_{n+2}=2+\cfrac{1}{{x}_{n+1}^{2}}+\cfrac{1}{{x}_{n}^{2}},\qquad n\geq 1.$ 讨论该数列

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$\begin{Bmatrix} {x}_{n} \end{Bmatrix}$ 的敛散性.

证明: (来自 magic9901) 设

lim n \to + \infty ˉ ˉ ˉ ˉ ˉ ˉ ˉ ˉ x n = A, lim n \to + \infty ˉ ˉ ˉ ˉ ˉ ˉ ˉ ˉ x n = B

$\lim\limits_{\overline{n\to+\infty}}x_n=A,\quad\overline{\lim\limits_{n\to+\infty}}x_n=B$ 于是就有

2<A?B<

$2<A\leqslant B<\cfrac{5}{2}$ ，则由递推关系式:

x n + 2 = 2 + 1 x 2 n + 1 + 1 x 2 n

$x_{n+2}=2+\cfrac{1}{x_{n+1}^2}+\cfrac{1}{x_n^2}$ 就得到:

A = = ? = = lim n \to + \infty ˉ ˉ ˉ ˉ ˉ ˉ ˉ ˉ x n + 2 lim n \to + \infty ˉ ˉ ˉ ˉ ˉ ˉ ˉ ˉ (2 + 1 x 2 n + 1 + 1 x 2 n) 2 + lim n \to + \infty ˉ ˉ ˉ ˉ ˉ ˉ ˉ ˉ 1 x 2 n + 1 + lim n \to + \infty ˉ ˉ ˉ ˉ ˉ ˉ ˉ ˉ 1 x 2 n 2 + 1 lim n \to + \infty ˉ ˉ ˉ ˉ ˉ ˉ ˉ ˉ x 2 n + 1 + 1 lim n \to + \infty ˉ ˉ ˉ ˉ ˉ ˉ ˉ ˉ x 2 n 2 + 2 B 2 .

$\begin{eqnarray*}A&=&\lim\limits_{\overline{n\to+\infty}}x_{n+2}\\ &=&\lim\limits_{\overline{n\to+\infty}}\left(2+\cfrac{1}{x_{n+1}^2}+\cfrac{1}{x_n^2}\right)\\ &\geqslant&2+\lim\limits_{\overline{n\to+\infty}}\cfrac{1}{x_{n+1}^2}+\lim\limits_{\overline{n\to+\infty}}\cfrac{1}{x_n^2}\\ &=&2+\cfrac{1}{\overline{\lim\limits_{n\to+\infty}}x_{n+1}^2}+\cfrac{1}{\overline{\lim\limits_{n\to+\infty}}x_n^2}\\ &=&2+\cfrac{2}{B^2}.\end{eqnarray*}$ 同理可以得到: