[SPAM] RE: [obm-l] Seq

[Artur Costa Steiner <artur_steiner@xxxxxxxxx>]

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--- Francis Alves <fran_cisca_sb@xxxxxxxxxxx> wrote:
Primeiro,vamos mostrar que a_n >= para todo n.

Para n = 1, isto se segue da definicao. Se valer para
algum n, entao = a_(n+1)= 3
- 1/a_n >=  3 - 1/2 = 2,5 > 2.

(i) seja s_n = a_n - a_( n - 1), n >=2. Entao, s_1 = 3
- 1/2 - 2 = 1/2 > 0. Supondo-se s_n > 0, entao s_(n +
1) = a_( n +1) - a_n = 3 - 1/a_n - (3 - 1/a_(n - 1)) =
 1/a_(n - 1) - 1/a_n . Como  os termos sao positivos
e, pela hipotese indutiva, a_(n - 1) < a_n, temos que
s_(n +) = 1/a_(n - 1) - 1/a_n > 0, o que completa a
inducao e mostra que a sequencia eh crescente. 

(ii) temos a_1 = 2 < 3. Se a_n < 3, entao a_(n+1)= 3
- 1/a_n < 3 - 1/3  < 3, o que completa a inducao e
mostra que a_n < 3 para todo n. 

(iii) acabamos de concluir que a_n e crescente e
limitada superiormente, o que implica que convirja
para algum real a. Em virtude da formula recursiva, a
uma das  raizes de

a = 3 - 1/a => a^2 - 3a + 1 = 0, 

as quais sao (3 - raiz(5))/2 < 2 e (3 + raiz(5))/2 >
2. Como a_n eh crescente e a_1 = 2, temos a > a_1 > 2,
de modo que a 1a raiz nao serve. Assim, temos a = (3 +
raiz(5))/2.

Artur  

> Tem sim.   Na verdade, o enunciado correto é:
>  
>  
> Mostre que a sequencia definida pora_1=2a_(n+1)= 3
> -1/a_ni) é crescente;ii)a_n<3 para todo n;iii) é
> convergente;iv) calcule seu limite.Fran ;-)
>  
> 
> 


      
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