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

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Let Fn be the nth Fibonacci number where F1 = F2 =1,  and Fn+1 = Fn+ Fn-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 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
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Next page : A second take on Taylor seriesLast modified: Mar 14, 2019 @ 21:42