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Re: Fibonacci


When using the add-on package Calculus`Limit` evaluation of 
Limit[Fibonacci[n]/Fibonacci[n-1], n -> Infinity] returns 1 rather than
GoldenRatio.   So, bug it is.  I will pass it on to the wri bugs
folks...


In article <6nsiqc$f39 at smc.vnet.net>, BobHanlon at aol.com wrote:

> 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?

-- 
David Reiss
dreissNOSPAM at nospam.earthlink.net
http://home.earthlink.net/~dreiss
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