I'm interesting to know about boundedness of the following sequence $ \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}$ with $F_n$ is a Fibonacci sequence such that I have tried to give its bound , we have $ \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}\geq \dfrac{F_n+F_{n-1}}{F_n}\geq 1+\dfrac{F_{n-1}}{F_n} $ using limit on both sides of that inequality we come up to $\lim_{n\to \infty} \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}\geq \lim_{n\to \infty} \dfrac{F_n+F_{n-1}}{F_n}\geq \lim_{n\to \infty} \big(1+\dfrac{F_{n-1}}{F_n}) =1+\dfrac{1}{\phi}$ but this it doesn't give any validity about boundedness of the titled sequence ? any help ?
2026-03-29 05:44:14.1774763054
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Is this $ \dfrac{F_n+F_{n-1}}{F_n-F_{n-1}}$ a bounded sequence with $F_n$ is a Fibonacci sequence?
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We have $$ \dfrac{F_{n}}{F_{n-1}} \to \phi $$ and so $$ \frac{F_n + F_{n-1}}{F_n - F_{n-1}} = \frac{\dfrac{F_{n}}{F_{n-1}}+1}{\dfrac{F_{n}}{F_{n-1}}-1} \to \frac{\phi+1}{\phi-1} =(\phi+1)\phi=\phi^2+\phi=2\phi+1 $$ Thus the sequence converges and so is bounded.
Alternatively, and simpler:
We have $ 0 < \frac{x+1}{x-1} < 5 $ for $x > \frac{3}{2}$. Now $\frac{F_{n}}{F_{n-1}} > \frac{3}{2}$ for $n \ge 5$; this follows easily by induction.
Note that $$ \frac{F_n + F_{n-1}}{F_n - F_{n-1}} = \frac{F_{n+1}}{F_{n-2}}. $$ We find that $$ \lim_{n \to \infty} \frac{F_n + F_{n-1}}{F_n - F_{n-1}} = \left(\lim_{n \to \infty} \frac{F_{n+1}}{F_n}\right) \cdot \left(\lim_{n \to \infty} \frac{F_{n}}{F_{n-1}}\right) \cdot \left(\lim_{n \to \infty} \frac{F_{n-1}}{F_{n-2}}\right) = \phi^3 = 2 \phi + 1, $$ where $\phi = \frac 12(1 + \sqrt{5})$. Because this sequence has a limit, it must be bounded.