Show that $\lim_{n\to\infty}\frac{1}{n}\log(F_{n})=\log(\varphi)$, where $(F_{n})$ is the Fibonacci sequence and $\varphi$ is the golden ratio

Im calculating the topological entropy of a certain dynamical system. In this calculation I encounter the limit $$\lim_{n\to\infty}\frac{1}{n}\log(F_{n}),$$ where $$(F_{n})$$ is the Fibonacci sequence ($$F_{1}:=1$$, $$F_{2}:=1$$, $$F_{n+2}:=F_{n+1}+F_{n}$$) and $$\varphi$$ is the golden ratio ($$=1.6180\ldots$$). How do I show that this limit converges to $$\log(\varphi)$$?

• Oct 30 '19 at 12:05

Note that $$\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\varphi$$ is a well known result. We also have that $$\lim_{n\to\infty}F_n^{1/n}=\varphi$$ because of the fact that $$\lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$ when the latter limit exists. So we can say that $$\lim_{n\to\infty}\frac{\log{(F_n)}}n=\lim_{n\to\infty}\log{\left(F_n^{1/n}\right)}=\log{\left(\lim_{n\to\infty}F_n^{1/n}\right)}=\log{(\varphi)}$$ because $$\log{(x)}$$ is continuous when $$x=\varphi$$.
Since $$\lim_{n\to\infty}\dfrac{F_{n+1}}{F_n}=\varphi$$, we have by continuity
$$\lim_{n\to \infty }\bigl(\log F_{n+1} -\log F_n\bigr)=\log\varphi.$$
Now use Cesàro's theorem and the fact that the series $$\bigl(\log F_{n+1} -\log F_n\bigr)$$ is a telescoping series.
$$F_n=\frac{a^n-b^n}{\sqrt{5}}, a= \frac{1 + \sqrt{5}}{2}, b=\frac{1-\sqrt{5}}{2}, a>|b|$$ So when $$n$$ is large $$\frac{a^n-2b^n}{\sqrt{5}}< F_n <\frac{2 a^n}{\sqrt{5}}$$ Next, $$\ln a +\frac{1}{n} \ln \frac{1}{\sqrt{5}}+ \ln[1-2 \left(\frac{b}{a}\right)^n] < \frac{1}{n} \ln (F_n) < \log a +\frac{1}{n} \ln \frac{2}{\sqrt{5}}.$$ As $$|b/a|<1$$ the limit $$n\rightarrow \infty$$ of both sides goes to $$\ln a.$$