In an infinite series of Fibonacci numbers, is this always true $$\frac{F_{n}}{F_{n+1}}>\frac{F_{n-1}}{F_n}$$? Can you make an argument that in an infinite convergent series, eventually that will be false?
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Or at least that the interval will tend to 0 so there can't be one rational number that will always be inside that interval? – Angela Oct 12 '16 at 17:12
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do you mean ? $$\frac{F_n}{F_{n+1}}>\frac{F_{n-1}}{F_n}$$ – Khosrotash Oct 12 '16 at 17:12
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That's what's there, right? – Angela Oct 12 '16 at 17:14
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$F_n+1 $ vs $F_{n+1}\ $,which one do you want to use ? – Khosrotash Oct 12 '16 at 17:16
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Oh yeah $F_{n+1}$ – Angela Oct 12 '16 at 17:17
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For the standard Fibonacci sequence: $F_0=F_1=1$, $F_2=2$, it's false at the very beginning. – Bernard Oct 12 '16 at 17:27
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The ratio of two consecutive Fibonacci numbers is a convergent of the continued fraction $[1;1,1,1,1,\ldots]$ of $\frac{1+\sqrt{5}}{2}$. The sequence of convergents is never monotonic, but the sequence of convergents with even (or odd) indices always is. – Jack D'Aurizio Oct 12 '16 at 17:36
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$$ F_{n+1}^2- F_n F_{n+2} = (-1)^n $$ is straightforward to prove by induction, too. – Jack D'Aurizio Oct 12 '16 at 17:37
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See also: Fibonacci ratio - inequality proof: $\frac{f_{2k}}{f_{2k-1}} < \frac{f_{2k+1}}{f_{2k}}$ and How do I prove $\frac{f_{2n+1}}{f_{2n}}>\frac{f_{2n}}{f_{2n-1}}$? – Martin Sleziak Dec 04 '20 at 22:16
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hint :We know **$\color{red} {F_{n+1}=F_n+F_{n-1}}\\ \to \color{red} {F_{n}=F_{n+1}-F_{n-1}}$ $$\frac{F_{n+1}-F_{n-1}}{F_{n+1}}>\frac{F_{n-1}}{F_{n+1}-F_{n-1}} $$ or see this $$\begin{bmatrix}f_{n+1} & f_n \\f_n & f_{n-1} \end{bmatrix}=\begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}^n \to \det(\begin{bmatrix}f_{n+1} & f_n \\f_n & f_{n-1} \end{bmatrix})=\det(\begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}^n)\\f_{n+1}f_{n-1}-f_n.f_n=(-1)^n\\n=2k \to f_{n+1}f_{n-1}-f_n.f_n=+1 \\f_{n+1}f_{n-1}>f_n.f_n\\ \frac{f_{n+1}}{f_{n}}>\frac{f_{n}}{f_{n-1}}\\ n=2k+1 \to f_{n+1}f_{n-1}-f_n.f_n=-1 \to f_{n+1}f_{n-1}<f_n.f_n \\ \to \frac{f_{n+1}}{f_{n}}<\frac{f_{n}}{f_{n-1}} \to \\ \color{red} {\frac{f_{n}}{f_{n+1}}>\frac{f_{n-1}}{f_{n}}}$$ this is your case when $n$ is odd
Martin Sleziak
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Khosrotash
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If you look at the fractions $2/3$, $3/5$ and $5/8$ you'll see that the inequality you want isn't always true. In fact it alternates between being true and false. In the limit the sequence (not series) is the golden ratio. Every other term is bigger/smaller; the difference goes to zero.
You can find a proof on the wikipedia page https://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio
Ethan Bolker
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I suspected this but thanks for confirming that I can use this in a proof. – Angela Oct 12 '16 at 17:54
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$$\frac{F_{n}}{F_{n+1}}>\frac{F_{n-1}}{F_n}$$ is equivalent to $$F_{n}^{2} - F_{n-1} F_{n+1} > 0.$$ Using \begin{align} 5 \, F_{n}^{2} &= L_{2n} - 2 (-1)^{n} \\ 5 \, F_{n-1}F_{n+1} &= L_{2n} + 3 (-1)^{n}, \end{align} where $L_{n}$ are the Lucas numbers, then $$-5 (-1)^n > 0$$ is the result. Only the odd values of $n$ make the given inequality true, ie $$\frac{F_{2n+1}}{F_{2n+2}}>\frac{F_{2n}}{F_{2n+1}}$$
Leucippus
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