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MathGroup Archive 1998

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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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