Cómo probar esta desigualdad para$a_{n}<5$

Question

Permitir que$a_{1}=1,a_{2}=2$ y estar relacionado por la recurrencia$$a_{n+2}=\dfrac{a^2_{n+1}(1+a_{n})+(1+2a_{n})a_{n+1}-a^2_{n}}{a_{n+1}+a^2_{n}+a_{n}+1}$ $

para $n\in N.$

¿Cómo puedo mostrar que$a_{n}<5,n\in N$?

mi idea: Creo que tenemos$$a_{n+2}=pa_{n+1}+qa_{n}$ $

porque tenemos$$a_{n+2}a_{n+1}+a_{n+2}a^2_{n}+a_{n+2}a_{n}+a_{n+2}=a^2_{n+1}+a^2_{n+1}a_{n}+a_{n+1}+2a_{n}a_{n+1}-a^2_{n}\cdots (1)$ $ entonces$$a_{n+1}a_{n}+a_{n+1}a^2_{n-1}+a_{n+1}a_{n-1}+a_{n+1}=a^2_{n}+a^2_{n}a_{n-1}+a_{n}+2a_{n-1}a_{n}-a^2_{n-1}\cdots (2)$ $ (1) (2)

Answer 1

Comenzaremos con la forma simplificada de la relación de recurrencia de @ Did: (con errores tipográficos corregidos):$$b_{n+2}=\dfrac{b_nb_{n+1}^2}{b_n^2-b_n+b_{n+1}}\quad (n=1,2,...)\quad\text{with}\quad b_1=2\text{ and }b_2=3.$$Write $ c_n: = \ dfrac {b_ {n 1} b_n ^ 2} {b_nb_ {n 1}} ;, (n = 1,2, ...). $ Then substituting for $$ in the corresponding expression for $$ and simplifying gives$$c_{n+1}=c_n\quad (n=1,2,...).$$Therefore $$ forall $$; that is $$ or$$b_{n+1}=\dfrac{6b_n^2}{7b_n-6}.$$Now the map$$ satisfies $$ and $$ for $$ (the minimum is $$ at $$). Since $$ is in this range, so are all $ $. Thus $c_n = c_1 = \ frac76$ for all $ n% #% #% (b_ {n 1} b_n ^ 2) / b_nb_ {n 1} = \ frac76,% #% # %% ff (x) = 6% #% #% 1 f (x) #% # %% #% #% f: x \ mapsto 6x ^ 2 / (7x-6) <6% #% #% 1 <x <6% #% #% \ frac {144} {49} 2)% #% #% b_n% #% #% b_n <6% #% #% n.