Let $\Phi$ be the golden ratio and $F_n$ be the usual Fibonacci numbers. How can I derive the following formula?

$$ \Phi = \lim_{n\rightarrow \infty} \sqrt[n]{F_n} $$

I know the usual relation $$ \Phi = \lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n} \quad , $$ and Wikipedia tells me that $$ \Phi^a = \lim_{n\rightarrow \infty} \frac{F_{n+a}}{F_n} \quad . $$

My first idea was to set $a = n$, which gives $$ \Phi = \lim_{n\rightarrow \infty} \sqrt[n]\frac{F_{n+n}}{F_n} \quad , $$

EDIT: We can also do $$ \Phi = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{F_{n+n}}{F_n}\frac{F_n}{F_n}} \quad , $$ but I am totally stuck here...

  • 7
    $\begingroup$ One way following what you know. $\endgroup$
    – OR.
    Mar 27, 2014 at 13:01
  • $\begingroup$ The work you have done so far suggests that as n gets large $F_{2n}$ approaches $F_n^2$ $\endgroup$
    – kleineg
    Mar 27, 2014 at 13:05
  • $\begingroup$ If you are willing to use Wikipedia, the same page also has some formulas saying, more or less, $F_n \approx\phi^n/\sqrt{5}$. Therefore $\sqrt[n]{F_n} \approx \phi / \sqrt[(2n)]{5}$. $\endgroup$ Mar 27, 2014 at 13:05
  • $\begingroup$ @kleinig: I cannot see this. Can you elaborate? One of my ideas was to multiply the term under the root by $F_n/F_n$ - when I then apply your hint, the result follows. $\endgroup$
    – mort
    Mar 27, 2014 at 13:12
  • $\begingroup$ @mort Sorry, that was less of a hint and more of an observation. When you set $a = n$ you got $\frac{F_{2n}}{F_n}$ in the nth root. And (once you prove it) we have the same equation with $F_n$ in the nth root. That suggests that $\frac{F_{2n}}{F_n} = F_n$ as n goes to infinity. $\endgroup$
    – kleineg
    Mar 27, 2014 at 13:27

5 Answers 5


It is a standard result that $F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}$.

Then $\lim_{n\rightarrow \infty} \sqrt[n]{F_n} = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}} = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{\phi^n}{\sqrt{5}}} = \lim_{n\rightarrow \infty} \frac{\phi}{\sqrt[n]{\sqrt{5}}} = \phi$ .

  • 1
    $\begingroup$ Ok, I actually understand this! $\endgroup$
    – mort
    Mar 27, 2014 at 13:11

We have

$$F_n=\frac1{\sqrt5}\left(\underbrace{\frac{1+\sqrt5}{2}}_{=:\alpha}\right)^n-\frac1{\sqrt5}\left(\underbrace{\frac{1-\sqrt5}{2}}_{=:\beta}\right)^n$$ and since $|\beta|<|\alpha|$ then $$|\beta|^n=_\infty o(|\alpha|^n)$$ hence $$\sqrt[n]{F_n}\sim_\infty \alpha=:\Phi$$


The process is to look at the approximation that $F_n = \Phi^n/\sqrt{5}$. The n'th root of this is $\sqrt[n]{F_n} = \Phi / 5^{1/2n}$.

The denominator approaches unity.


Let $\Phi_n=\frac{F_{n+1}}{F_n}$ and $R_n=\sqrt[n]{F_n}$, then we have $$R_n^n=F_n=F_{n-1}\Phi_{n-1}=R_{n-1}^{n-1}\Phi_{n-1}$$ Take $n\to\infty$, $$R^n=R^{n-1}\Phi$$ Hence $R=\lim_{n\to\infty}R_n=\Phi$.

  • $\begingroup$ Sorry, it is not clear to me why you could make $n\to \infty$ on the subindex but not on the uperindex. Could you explain? $\endgroup$ Mar 6, 2021 at 1:35

We know $$\lim_{n \to \infty} \frac{F_{n+1}}{F_n}=\Phi$$

by continuity of logarithm function, this implies

$$\lim_{n \to \infty} (log{F_{n+1}}-log{F_n})=log\Phi$$

By definition, $\forall{\epsilon \gt 0}, \exists{N\in {\mathbb N}}$ such that $n \ge N $ implies $|log{F_{n+1}}-log{F_n}-log\Phi|\lt \epsilon_1$

So by triangle inequality, $$|(\sum_{k=N}^{n}\frac{1}{n-N+1}(logF_{k+1}-logF_k))-log\Phi|\le \frac{1}{n-N+1}\sum_{k=N}^{n}|logF_{k+1}-logF_k-log\Phi| \lt \epsilon_1 $$

Let $$\sum_{k=1}^{N-1} |logF_{k+1}-logF_k-log\Phi|=A$$

Then $$|\sum_{k=1}^{n}(logF_{k+1}-logF_k-log\Phi)| \le \sum_{k=1}^{n} |logF_{k+1}-logF_k-log\Phi|\;,$$

which can be split into $$\sum_{k=1}^{N-1} |logF_{k+1}-logF_k-log\Phi|+\sum_{k=N}^{n} |logF_{k+1}-logF_k-log\Phi|$$

Hence, we have $$\frac{1}{n-N+1}(\sum_{k=1}^{N-1} |logF_{k+1}-logF_k-\Phi|+\sum_{k=N}^{n} |logF_{k+1}-logF_k-log\Phi|)\lt \frac{A}{n-N+1}+\epsilon_1$$

Now, fix $N=N_0$. Since A is also fixed, $A \gt 0, \forall{\epsilon_2 \gt 0}, \exists{N_1 \in{\mathbb N}}$ such that $n \ge N_1 \;\text{implies} \; (n-N_0+1)\epsilon_2\lt A, \;\text{so}\; \dfrac{A}{n-N_0+1}\lt \epsilon_2.$

Hence, $\forall{n \ge N_1},$

$$|(\sum_{k=1}^{n}\frac{1}{n}(logF_{k+1}-logF_k))-log\Phi|\lt|\sum_{k=1}^{n}\frac{1}{n-N_0+1}(logF_{k+1}-logF_k-log\Phi)|\lt \epsilon_1+\epsilon_2$$

$$|\frac{1}{n}logF_{n+1}-log\Phi|=|(\sum_{k=1}^{n}\frac{1}{n}(logF_{k+1}-logF_k))-log\Phi|\lt \varepsilon_1$$

Therefore, by definition,

$$lim_{n \to \infty}\frac{1}{n}logF_{n+1}=lim_{n \to \infty}log\sqrt[n]{F_{n+1}}=log\Phi$$

By the continuity of the exponential function $e^x$, we get $$lim_{n\to \infty} \sqrt[n]{F_{n+1}}=\Phi$$

Hence proven.

Remark: "Introduction to Calculus and Analysis I" by Richard Courant and Fritz John,page 114,SECTION 1.7, exercise $^*10$: "If $a_n \gt 0, \;\text{and} \; lim_{n\to \infty}\dfrac{a_{n+1}}{a_n}=L,\;\text{then}\;lim_{n\to \infty}\sqrt[n]{a_n}=L."$


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