Questao de sequencias

["Luis Lopes" <llopes@xxxxxxxxxxxxx>]

Sauda,c~oes,

Segue a solu,c~ao do prof. Rousseau para a
quest~ao.

[ ]'s
Lu'is


Dear Luis:

   Perhaps I have missed something, but it seems to me
that the  sequence problem is quite easy (once you forget
about the strange expression on the right and simply try
to prove an upper bound).  Write a_i = (-1)^{i-1}b_i for
i \geq 1.  Then the recurrence gives
b_{i+1}/b_i  = (2 - (b_{i-1}/b_i)^2), so |b_i/b_{i-1}| > 1
implies b_{i+1}/b_i > 1.  Thus inductively we find that
(b_1, b_2, b_3, ...) is an increasing positive sequence.
Then
1/a_0  + 1/a_1 + \cdots + 1/a_k = 1 + 1/a - 1/b_2 + 1/b_3 - \cdots
(-1)^{k-1}/b_k
= (a+1)/a - (1/b_2 - 1/b_3) - (1/b_4 - 1/b_5) - \cdots < (a+1)/a.
(The last term is either - (1/a_{k-1} - 1/a_k) or - 1/a_k; in either
case it is negative.)   Then the sum is less than (a + 2 - \sqrt{a^2-4})/2
since the latter satisfies
        1 + \frac{a - \sqrt{a^2-4}}{2} = 1 + \frac{2}{a + \sqrt{a^2-4}}
          > 1 + 1/a.
Thus we improve the stated result by a factor of 2.  However the simple
bound (a+1)/a is even better, so the complicated expression seems
to have no role to play
There are some other interesting things going on here, but they don't
lead directly to a solution of the problem as posed.  Namely, if we
write  r_n = b_n/b_{n-1} (so b_n = b_0 r_1 r_2 \cdots r_n), then
r_{n+1} = 2 - 1/r_n^2.  Then with r_1 > 1 the sequence (r_n)
converges to the golden ratio \phi =  (1 + \sqrt{5})/2, so
asymptotically the infinite series \sum_k 1/a_k behaves like a
geometric series with common ratio - 1/\phi.

Cecil

> -----Mensagem Original-----
> De: Sistema ELITE de Ensino - Unidade Belém
> Para: obm-l@mat.puc-rio.brEnviada em: Terça-feira, 26 de Setembro de
> 2000 16:13Assunto: Questao de sequencias
>  Não consigo fazer esta questão de seqüências do banco da IMO de 1996.
>
> Questão: Seja a > 2 um número dado, e defini-se recursivamente a(0) =
> 1, a(1) = a,
>
> a(n+1) = ((a(n-1)/a(n))^2 - 2)a(n).
>
> Mostre que para todos os inteiros k > 0 tem-se
>
> 1/a(0) + 1/a(1) + 1/a(2) + ... + 1/a(k) < 2 + a - (a^2 - 4)^(1/2)
>
> Obs: Evidentemente, a(i) indica: a índice i.