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        riddles >> putnam exam (pure math) >> Fibonacci convergence...
        
(Message started by: Michael_Dagg on Jun 23rd, 2006, 8:01am)
Title: Fibonacci convergence...
Post by Michael_Dagg on Jun 23rd, 2006, 8:01am
Let F be Fibonacci. Discuss the convergence of

H = F_1 + sqrt(F_2 + sqrt(F_3 + sqrt(F_4 + ...))).
Title: Re: Fibonacci convergence...
Post by Barukh on Jun 25th, 2006, 4:32am
Here’s my attempt on the solution.

1. Define Hn = F1 + sqrt(F2 + sqrt(…+sqrt(Fn)…). Then {Hn} is monotonically increasing sequence.

2. Define Gn = 21 + sqrt(22 + sqrt(…+sqrt(2n)…). Because Fn < 2n, it follows that Fn < Gn.

3. Define also In = 21sqrt(22sqrt(…sqrt(2n)…), where every addition is substituted by a product. It is easy to see that Gn <= In.

4. But In = 2En, where En = sum i21-i. This last sequence has a finite limit (vis. 4).

5. Therefore, Hn < 16, and converges.

Of course, better estimates can be obtained, together with much more general results.
Title: Re: Fibonacci convergence...
Post by Barukh on Jun 25th, 2006, 11:47pm
A better estimate is obtained by using the following identity: (2n+1)2 = 22n+(2n+1+1), therefore

3 = (21+1) = sqrt(22+(22+1)) = sqrt(22+sqrt(24+(23+1))) = … = sqrt(22+sqrt(24+ sqrt(26+…))).

Now, because Fn < 22n-2, it follows H < 4.
Title: Re: Fibonacci convergence...
Post by towr on Jun 26th, 2006, 2:40am
In a similar way as above, I've tried using
(sqrt(2)n + 1)2 = 2n + 2 sqrt(2)n + 1 > 2n + sqrt(2)n+1 + 1
F(n) <= 2n-2 for n >= 2

2 = sqrt(2)0 + 1
> sqrt(20 + (sqrt(2)1 + 1))
> sqrt(20 + sqrt(21 + (sqrt(2)2 + 1)))
> sqrt(20 + sqrt(21 +  sqrt(22 + ... )))
>= sqrt(F(2) + sqrt(F(3) + ...))

1 + 2 = 3 > H

[edit]finally found where I went wrong..[/edit]


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