数列$egin{Bmatrix} {x}_{n} end{Bmatrix}$满足如下定义: $$a>0,quad b>0; qquad {x}_{1}=a,quad{x}_{2}=b ;qquad {x}_{n+2}=2+cfrac{1}{{x}_{n+1}^{2}}+cfrac{1}{{x}_{n}^{2}},qquad ngeq 1.$$ 讨论该数列 $egin{Bmatrix} {x}_{n} end{Bmatrix}$ 的敛散性.
证明: (来自 magic9901) 设 [limlimits_{overline{n o+infty}}x_n=A,quadoverline{limlimits_{n o+infty}}x_n=B] 于是就有 $2<Aleqslant B<cfrac{5}{2}$，则由递推关系式: [x_{n+2}=2+cfrac{1}{x_{n+1}^2}+cfrac{1}{x_n^2}] 就得到: $$egin{eqnarray*}A&=&limlimits_{overline{n o+infty}}x_{n+2}\ &=&limlimits_{overline{n o+infty}}left(2+cfrac{1}{x_{n+1}^2}+cfrac{1}{x_n^2} ight)\ &geqslant&2+limlimits_{overline{n o+infty}}cfrac{1}{x_{n+1}^2}+limlimits_{overline{n o+infty}}cfrac{1}{x_n^2}\ &=&2+cfrac{1}{overline{limlimits_{n o+infty}}x_{n+1}^2}+cfrac{1}{overline{limlimits_{n o+infty}}x_n^2}\ &=&2+cfrac{2}{B^2}.end{eqnarray*}$$ 同理可以得到: [Bleqslant 2+cfrac{2}{A^2}] 因此就有: [0leqslant B-Aleqslant cfrac{2}{A^2}-cfrac{2}{B^2}=cfrac{2(A+B)}{A^2B^2}(B-A)] 注意到: [cfrac{128}{625}<cfrac{2(A+B)}{A^2B^2}<cfrac{5}{8}] 则得到: $A=B$, 所以就有: [limlimits_{n o+infty}x_n=A=B=xinsex{2,cfrac{5}{2}}.]