Convergence of Binet's formula expression for Fibonacci

Question

Let $\displaystyle \phi = \frac{1+\sqrt{5}}{2}$ and $\displaystyle \psi = \frac{1-\sqrt{5}}{2}$. Consider the Fibonacci sequence defined by:

$$ \displaystyle a_n = \frac{\phi^n - \psi^n}{\sqrt{5}} $$

Evaluate $\displaystyle \lim_{n \to \infty} \frac{\log(a_n)}{n}$.

My feeling is that I should work with the sequence $\displaystyle a_n^{1/n}$ and then $\log$ the result but I may be completely off. Any suggestions on how to go about this?

Answer 1

For large $n$, $\psi^n$ goes to $0$, so $a_n\approx\frac{\phi^n}{\sqrt{5}}$ for large $n$. Hence $$\displaystyle \lim_{n \to \infty} \frac{\log(a_n)}{n}=\lim_{n \to \infty} \frac{\log\left(\frac{\phi^n}{\sqrt5}\right)}{n}=\log(\phi) $$

Answer 2

Note that $\phi = -1 / \psi$. Then $$a_n = \frac {\phi ^ {2n} + (-1)^{n + 1}} {\phi ^ n \sqrt 5} =\frac {\phi^n} {\sqrt 5} + o (1).$$ Then $$\log a_n = n \log \phi + \varepsilon_n$$ where $\varepsilon_n \to 0$. Hence, $$\frac {\log a_n} {n} \to \log \phi = 0.481211825\ldots$$