I am trying to find $n\in \Bbb{N}$ such that $$\forall m\geq n\quad \lvert \frac{F_m}{F_{m-1}} – \phi \rvert < \frac{1}{100}$$ where $F_n$ is the n-th term of the Fibonacci sequence and $\phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. I know and have proved that $\lim_{n\to\infty} \frac{F_n}{F_{n-1}} = \phi$ using Binet's formula, however this looks like a formal limit. I would have to express $\lvert \frac{F_m}{F_{m-1}} – \phi \rvert < f(m)$ so that I could find $m$ for $f(m) = \frac{1}{100}$.
I have managed to express $\frac{F_m}{F_{m-1}}$ as $\frac{1}{2} \frac{(1+\sqrt{5})^{n+1} – (1-\sqrt{5})^{n+1}}{(1+\sqrt{5})^{n} – (1-\sqrt{5})^{n}}$ using Binet's formula, however this didn't bring me far.
Through numerical calculations I have found that :
\begin{array}{c|c}
m & \frac{F_m}{F_{m-1}} \\\hline
1 & 1 \\\hline
2 & 2 \\\hline
3 & 1.5 \\\hline
4 & 1.66667 \\\hline
5 & 1.6 \\\hline
6 & 1.6250 \\\hline
7 & 1.6154 \\\hline
8 & 1.6190 \\\hline
9 & 1.6176 \\\hline
10 & 1.6182
\end{array}
However, I couldn't find a proof for this fact. So my question, is there any way to do a formal proof of the convergence of the slope of the Fibonacci sequence ?
Best Answer
Hint
If you look here
$$\frac {F_{n + 1} } {F_n}= \phi + \dfrac {\sqrt 5} {\alpha^n - 1}\quad \text{where} \qquad \alpha=\dfrac {\phi} {\hat \phi}=\dfrac {1 + \sqrt 5} {1 - \sqrt 5}$$ So, looking for $n$ such that $$\left|\frac{F_{n+1}}{F_n}-\phi \right|\leq \epsilon \implies \left|\alpha^n-1\right| \geq \frac{\sqrt 5}\epsilon $$ It looks simple (taka care that $\alpha <0$).