Proof that ratios of Fibonacci successive numbers tend to the Golden ratio

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Let F_n be the n^th Fibonacci number where F₁ = F₂ =1, and F_n₊₁ = F_n+ F_n_-1.

We should prove that

$\mathop {\lim }\limits_{k \to \infty } \frac{{\;{F_{k + 1}}\;}}{{{F_k}}} = \frac{{1 + \sqrt 5 }}{2}$

i.e. the golden ratio. We have that

$\begin{gathered} \mathop {\lim }\limits_{k \to \infty } \frac{{\;{F_{k + 1}}\;}}{{{F_k}}} = \mathop {\lim }\limits_{k \to \infty } \frac{{\;{F_k} + {F_{k - 1}}\;}}{{{F_k}}} \hfill \\ \quad = \mathop {\lim }\limits_{k \to \infty } \left( {1 + \frac{{\;{F_{k - 1}}\;}}{{{F_k}}}} \right) = 1 + \mathop {\lim }\limits_{k \to \infty } \frac{{\;{F_{k - 1}}\;}}{{{F_k}}} \hfill \\ \quad = 1 + \mathop {\lim }\limits_{k \to \infty } \frac{{\;{F_k}\;}}{{{F_{k + 1}}}} = 1 + \mathop {\lim }\limits_{k \to \infty } \left( {1/(\;{F_k}/{F_{k + 1}})} \right) \hfill \\ \end{gathered}$

or, if

$x = \mathop {\lim }\limits_{k \to \infty } \frac{{\;{F_{k + 1}}\;}}{{{F_k}}}$

then

$x = 1 + \frac{1}{x}$

and then you solve this for x. You will get a quadratic that has a negative and a positive solution, but since all terms are positive, the positive root must be the right one, and as you can verify it is indeed the golden ratio.

Interestingly enough we never used the starting numbers in proving this. If one is negative we might get a sequence of negative numbers, but the result will still hold true. The only exception is if both the starting numbers are 0.

Up a level : Power Series
Previous page : Some interesting examples based on the Fibonacci series
Next page : A second take on Taylor series

Last modified: Dec 28, 2023 @ 09:46