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?