3
votes
3answers
76 views

Proof by induction: $n$th Fibonacci number is at most $ 2^n$

I'm trying to find the proof by induction of the following claim: For all $n\in\mathbb N$, $\operatorname{fibonacci}(n) \le 2^n$ My Proof: Base case: $n = 1$ $\operatorname{fibonacci}(1) \le 2^ 1$ ...
1
vote
2answers
69 views

How to establish this inequality without using induction?

Given the Fibonacci sequence $a_1 = 1$, $a_2 = 2$, $\ldots$, $a_{n+1} = a_n + a_{n-1} $ for $n \geq 2$, how to derive, without using induction, the inequality $$ a_n < (\frac{1+\sqrt{5}}{2})^n $$ ...
0
votes
1answer
28 views

Inequality of Lucas Numbers

Can it be shown that \begin{align} \frac{1}{\ln(1+L_{n}) -1} \geq \frac{L_{n}}{(L_{n}-1)(e^{L_{n}}-1)} \end{align} where $L_{n}$ is the $n^{th}$ Lucas number. Show results in full detail.
1
vote
1answer
28 views

Prove that $F_{\sum_{i=1}^ka_i}\geq \prod_{i=1}^kF_{a_i}\forall a_i,k \geq 1$

Is there an elegant way to do this? I don't think it's particularly difficult, since $F_n \sim \frac{\phi^n}{\sqrt{5}}$, so we expect that $F_{\sum_{i=1}^ka_i} \sim ...
1
vote
1answer
134 views

How to solve for the $n$-th Fibonacci number that is greater than or equal to $N$?

The general formula for the $n$-th Fibonacci number is: $$\frac{\phi^n - (1 - \phi)^n}{\sqrt{5}}$$ where $$\phi = \frac{1 + \sqrt{5}}{2}$$ Given $N$, is there a way to solve for $n$ in this ...
7
votes
1answer
573 views

Another way to go about proving the limit of Fibonacci's sequence quotient.

It is not difficult to inductively prove that $$\eqalign{ & \phi = \phi + 0 \cr & {\phi ^2} = \phi + 1 \cr & {\phi ^3} = 2\phi + 1 \cr & {\phi ^4} = 3\phi + 2 ...