What is the limit of the sequence: $n$-th root of the $n$-th Fibonacci number?

Question

My computer can not calculate numbers large enough for this.

If you take the $n$-th Fibonacci number $F_n$ and raise it to the $1/n$-th term, where does the sequence $F_n^{1/n}$ tend to?

Examples:

• $13^{1/7}$

• $21^{1/8}$

Answer 1

$F_n$ is the integer closest to $\frac{\varphi^n}{\sqrt5}$, where $\varphi=\frac{1}2\left(1+\sqrt5\right)$. Its $n$-th root is therefore approximately $\frac{\varphi}{5^{1/(2n)}}$, which approaches $\varphi$ as $n$ increases.

Answer 2

There are many ways to do this, but perhaps the easiest is a modified form of Binet's formula:

$$F_n=\left[\frac{\phi^n}{\sqrt{5}}\right],$$

where the brackets are the nearest integer function. You can squeeze the above between: $\frac{\phi^n}{\sqrt{5}}\pm 1$. Now confirm that:

$$(\frac{1}{\sqrt{5}}\phi^n+1)^{1/n}=\phi\exp\left(\frac{1}{n}\ln(\frac{1}{\sqrt{5}}+1/\phi^n)\right)\rightarrow \phi,$$

so by the squeeze theorem $F_n^{1/n}\rightarrow\phi$.

Answer 3

THEOREM:-$F_n = \frac{a^n-b^n}{a-b}$ where $a$ and $b$ are the roots for the polynomial $x^2-x-1 $.

PROOF:- Since $\frac{a^1-b^1}{a-b} = 1$ and $\frac{a^2-b^2}{a-b} = 1$, assume it work for $ F_n$ and $ F_{n+1 } $ then $a^{n+2}-a^{n+1}-a^n-b^{n+2}+b^{n+1}+ b^n =0 $ (why? factor $a^n $ and $b^n$ then notice it is the same polynomial as $x^2-x-1 $ ) which proves it work for $F_{n+2}$ by Induction ( rearrange the terms so ${a^{n+2}-b^{n+2}} ={a^n-b^n}+ {a^{n+1}-b^{n+1}}$ and divide both side by $ a-b $) Therefore, by induction, the formula holds for all positive integers $n$.

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Let $a>b$ $$ a\sqrt[n]{1-\left|\frac{b}{n} \right|^n}< \sqrt[n]{a^n - b^n }< a\sqrt[n]{1+\left|\frac{b}{n} \right|^n}$$

since $|a|> |b|$ then $\lim\limits_{n \to \infty}\left|\frac{b}{n} \right|^n =0$ then by squeeze theorem $\lim\limits_{n \to \infty}\sqrt[n]{a^n - b^n } =a$

$$\lim\limits_{n \to \infty}\frac{\sqrt[n]{a^n - b^n }}{\sqrt[n]{a-b}} =a$$