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Re: Fibonacci
*To*: mathgroup at smc.vnet.net
*Subject*: [mg13104] Re: [mg13052] Fibonacci
*From*: BobHanlon at aol.com
*Date*: Tue, 7 Jul 1998 03:44:09 -0400
*Sender*: owner-wri-mathgroup at wolfram.com
On a Macintosh running Mathematica 3.0.1, the limit returns unevaluated:
Limit[Fibonacci[n]/Fibonacci[n-1], n -> Infinity]
Limit[Fibonacci[n]/Fibonacci[-1 + n], n -> Infinity]
fib[n_] := (((1+Sqrt[5])/2)^n - (-((1+Sqrt[5])/2))^-n)/Sqrt[5]
Table[fib[n], {n, 10}]//FullSimplify
{1,1,2,3,5,8,13,21,34,55}
Limit[fib[n]/fib[n-1], n -> Infinity]
Limit[(-(1/2*(-1 - Sqrt[5]))^(-n) + (1/2*(1 + Sqrt[5]))^n)/
(-(1/2*(-1 - Sqrt[5]))^(1 - n) +
(1/2*(1 + Sqrt[5]))^(-1 + n)), n -> Infinity]
However, since
fib[n]/fib[n-1] == (fib[n-1] + fib[n-2])/fib[n-1] == 1 +
fib[n-2]/fib[n-1]
Then, the limit must satisfy the equation (x == 1 + 1/x)
Select[x /. Solve[x == 1 + 1/x, x], #>0&]
{1/2*(1 + Sqrt[5])}
which is the Golden Ratio
%[[1]] == GoldenRatio // N
True
Bob Hanlon
In a message dated 7/4/98 8:01:56 PM, tobi.kamke at t-online.de wrote:
>I've a problem. I thought that Limit[Fibonacci[n]/Fibonacci[n-1], n ->
>Infinity] is GoldenRatio.
>
>Mathematica says 1.
>What's wrong?
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