Continuous Fibonacci number F(n)

$F_n=\frac{\phi^n-\cos(\pi n)\phi^{-n}}{\sqrt{5}},$ with $\phi$ being the golden ratio. Here $n$ can be also complex.

You can also rewrite the ratio as $$\frac{F_{n+1}}{F_n}=\phi\left(\frac{1+(-1)^{n+1}\phi^{-2(n+1)}}{1+(-1)^{n+1}\phi^{-2n}}\right),$$ where it easier to show that the ratio converges to $\phi$ and maybe you like it for calculations. But for the proof only I would rather prefer the other answers to be honest.

Since you are using Mathematica, judging by the code in the OP, you can easily obtain the actual expression used by Mathematica for evaluating Fibonacci numbers of non-integer index:

FunctionExpand[Fibonacci[n]]
   ((1/2 (1 + Sqrt[5]))^n - (2/(1 + Sqrt[5]))^n Cos[n π])/Sqrt[5]

In more conventional notation ($\phi$ is the golden ratio, as usual):

$$F_n=\frac{\phi ^n-\phi ^{-n} \cos (\pi n)}{\sqrt{5}}$$

(This is also the expression given in the Wolfram Functions site and CostaZach's answer.)

This of course doesn't imply that this is the only way to define the Fibonacci numbers for non-integer index. For instance, this function uses

$$f_n=\frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n\sqrt{5}}$$

but unlike the previous formula, this becomes complex-valued in general for real non-integer argument.

$\DeclareMathOperator{\bbN}{\mathbb{N}}$$\newcommand{\digsum}{\text{digitsum}} \DeclareMathOperator{\bbO}{\mathbb{O}} \DeclareMathOperator{\bbE}{\mathbb{E}}$$\blacksquare$ Notation: Let us denote:

• $\bbO := $ the set of all odd natural numbe\rs
• $\bbE := $ the set of all even natural numbers

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Hint: Let $F_0 = 0$ and $F_1 = 1$. We can see the following: $$ F_{n + 1} = F_{n} + F_{n - 1} \quad \text{for } n \geqslant 1 $$ $$ \implies \frac{F_{n + 1}}{F_n} = 1 + \frac{F_{n-1}}{F_n} = 1 + \frac{1}{\frac{F_{n}}{F_{n - 1}}} = 1 + \frac{1}{1 + \frac{F_{n-2}}{F_{n - 1}}} = \cdots = 1 + \frac{1}{1 + \frac{1}{ 1 + \frac{1}{1 + \frac{1}{1 + \cdots (n\text{ times})}}}} $$

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Lemma 1: The sequence $\{ \xi_n\}_{n=1}^\infty := \left\{ \dfrac{F_{n + 1}}{F_n} \right\}_{n = 1}^\infty$ has the following properties:

• The even subsequence $\{ \xi_{2k}\}_{k=1}^\infty := \left\{ \dfrac{F_{2k + 1}}{F_{2k}} \right\}_{k = 1}^\infty$ increases.

• The odd subsequence $\{ \xi_{2k + 1}\}_{k=0}^\infty := \left\{ \dfrac{F_{2k + 2}}{F_{2k + 1}} \right\}_{k = 0}^\infty$ decreases.

Proof: The proof will be immediate from the relation $~F_{n}^2 + (-1)^n = F_{n + 1}F_{n - 1} ~$ for any $n \geqslant 1$ [prove it using induction].

Now we have that $$ \xi_{n+1} - \xi_n = \frac{F_{n + 1}}{F_n} - \frac{F_n}{F_{n - 1}} = \frac{F_{n +1}F_{n - 1} - F_n^2 }{F_n F_{n - 1}} = \frac{(-1)^n}{F_nF_{n - 1}} \begin{cases} >0 & \text{if } n \in \bbE\\ < 0 & \text{if } n \in \bbO \end{cases} $$ Thus we proved our claim. $\hspace{14cm} \blacksquare$

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Lemma 2: The sequence $\{ \xi_n\}_{n = 1}^\infty$ is bounded.

Proof: We know that $\{ F_n\}_{n = 0}^\infty$ is an increasing sequence. Note that $$ F_{n + 1} = F_{n} +F_{n - 1} \implies \xi_n = 1 + \frac{F_{n - 1}}{F_n} < 1 + 1 = 2 $$ And trivially $ 0 < \xi_n < 2 ~$ for all $n \in \bbN$

Thus we have that $$0 <\frac{F_{n+1}}{F_{n}} < {2} $$ Hence it's bounded. $\hspace{15cm} \blacksquare$

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Lemma 3: The following holds: $$ \lim_{n \to \infty} \frac{F_{n + 1}}{F_n} ~\text{ exists } $$ Proof: Follows from Lemma 1 and Lemma 2 $\hspace{7cm} \blacksquare$

So, let's observe that $$\lim\limits_{n \to \infty} \frac{F_{n + 1}}{F_n } = L = 1 + \frac{1}{1 + \frac{1}{ 1 + \frac{1}{1 + \frac{1}{1 + \cdots (\infty\text{ times})}}}} = \frac{1}{1 + L} $$ $$ \implies L^2 - L - 1 = 0 \implies L = \frac{1 + \sqrt{5}}{2} = \phi $$