On Fibonacci sequence

Question

Let $F_n$ and $F_m$ be two terms of Fibonacci sequence such that $m<n$.

Is there an upper bound $c<1$, such that $\dfrac{F_m}{F_n}<c$ ?

Answer 1

Let $F_n$ and $F_m$ be two terms of Fibonacci sequence such that $m<n$

Is there an upper bound $c<1$, such that $\dfrac{F_m}{F_n}<c$ ?

For large $m<n$ you have $c \geq \frac{\sqrt{5}+1}{2}$ 1

The statement is false as we can demonstrate by the data that: $\dfrac{F_m}{F_n}> c$ where $c=1$ and $m < n$. 2

Furthermore, we know that $c \geq \phi = \frac{\sqrt{5}+1}{2}$ 1

To prove this:

$$\lim_{(n,m) \to (\infty, \infty)} (\frac{F_m}{F_n}) \stackrel{m < n}{\implies} \lim_{(n,m) \to (\infty, \infty)} (\frac{F_m}{F_n})<1$$

Furthermore, here we have an Excel computation for some finite examples:

Answer 2

(Throughout this answer it is assumed that we don't consider the case $F_m=F_n=1$, i.e. we only look at the strictly increasing part of the Fibonacci numbers.)

Using the observation from Crostul here is a proof that there exists some $c<1$ s.t. $\frac{F_m}{F_n}<c$ for all $m<n$.

We use the fact that the limit of $\frac{F_{n+1}}{F_n}$ as $n$ tends to infinity is $\phi=\frac{\sqrt{5}+1}{2}>1$.

Using the definition of the limit, for any $\epsilon>0$ there is an $N_{\epsilon}$ s.t. $|\frac{F_{n+1}}{F_n}-\phi|<\epsilon$ for all $n>N_{\epsilon}$.

Choosing any $0<\epsilon<\phi-1$ we can see that, letting $\delta=\phi-1-\epsilon$, for all $n>N_{\epsilon}$, $\frac{F_{n+1}}{F_n}>1+\delta$ and so $\frac{F_n}{F_{n+1}}<\frac{1}{1+\delta}<1$

Now as $F_n$ is strictly increasing $\frac{F_m}{F_n}$ is maximised with $n=m+1$ and so if $m>N_{\epsilon}$, $\frac{F_m}{F_n}<\frac{1}{1+\delta}$.

There are finitely many $m$ with $m \leq N_{\epsilon}$ and so $M=\max(\{\frac{F_m}{F_{m+1}}, m \leq N_{\epsilon}\})$ exists.

As $F_n$ is increasing $M<1$. And so $c=\max(M,\frac{1}{1+\delta})$<1 and for all $m<n$, $\frac{F_m}{F_n}<c$

$\blacksquare$

Finding the minimal such $c$ should essentially just be an exercise in computation.

Additionally essentially the same argument shows that for any sequence $a_n$ the existance of a $c$ as described is equivalent to the sequence being strictly increasing and having $\liminf \frac{a_{n+1}}{a_n}>1$ (or combined $\inf(\{ \frac{a_{n+1}}{a_n}, n \in \Bbb{N}$})>1 ).