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The Fibonacci function I have constructed is:

f[n_] := f[n] = f[n - 1] + f[n - 2]

f[0] = 0; f[1] = 1;

\[ScriptF][n_] := \[ScriptF][n] = f[n] + f[n - 1]

Is there a way to take the limit of the ratios of

\[ScriptF][n_] := \[ScriptF][n] = f[n] + f[n - 1] / f[n_] := f[n] = f[n - 1] + f[n - 2]

as n tends to infinity?

Any help is greatly appreciated.

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2 Answers 2

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RSolve andLimit will do the equivalent:

Clear[f];
rsol = RSolve[{f[n] == f[n - 1] + f[n - 2], f[0] == 0, f[1] == 1}, f, n]
(*  {{f -> Function[{n}, Fibonacci[n]]}}  *)

Limit[f[n]/f[n - 1] /. First[rsol], n -> Infinity]
(*  1/2 (1 + Sqrt[5])  *)
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  • $\begingroup$ An equivalent solution: f = DifferenceRoot[Function[{f, n}, {f[n] == f[n - 1] + f[n - 2], f[0] == 0, f[1] == 1}]]; Limit[f[n + 1]/f[n] // FunctionExpand, n -> ∞] $\endgroup$ Commented Mar 27, 2017 at 11:40
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Or use the built in function Fibonnaci:

Limit[Fibonacci[n + 1]/Fibonacci[n], n -> Infinity]
(*  1/2 (1 + Sqrt[5])  *)
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