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 ?
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5Note that $$ \frac{F_n + F_{n-1}}{F_n - F_{n-1}} = \frac{F_{n+1}}{F_{n-2}}. $$ – Ben Grossmann Nov 13 '20 at 21:04
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This should help: https://math.stackexchange.com/q/132305/42969 – Martin R Nov 13 '20 at 21:05
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just prove something crude like $1\leq \dfrac{F_n}{F_{n-1}} < 2$ by induction, then you get an upper and lower bound. Anything stronger, you will need to delve into more properties of the Fibonacci sequence. – dezdichado Nov 13 '20 at 21:17
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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.
Ben Grossmann
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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.
lhf
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