2
$\begingroup$

Here is one of the exercices that I need to solve for my Analysis course:

Given the Fibonacci sequence where $a_0:=1$, $a_1:=1$, $a_n:=a_{n−1}+a_{n−2}$ and the Fibonacci ratio sequence $(b_n)$ defined as $$ b_n:=\frac{a_n}{a_{n-1}} $$

How can I show that the sequence $(b_{2n})$ is decreasing and the sequence $(b_{2n+1})$ is increasing?

I tried working with ex.: $b_{2n}-b_{2n-2}$ being $\le0$ or the ratio $\frac{b_{2n}}{b_{2n-2}}$ being $<1$, but I can't find the right path.

Any tips?

$\endgroup$

1 Answer 1

2
$\begingroup$

It is easy to show by induction that \begin{eqnarray*} a_{2n}^2-a_{2n+1}a_{2n-1}=1. \end{eqnarray*} Now create a couple of terms & use the recurrence relation \begin{eqnarray*} a_{2n}^2 +\color{blue}{a_{2n}a_{2n-1}-a_{2n}a_{2n-1}}-a_{2n+1}a_{2n-1}=1 \\ \color{red}{a_{2n}^2 +a_{2n}a_{2n-1}}-\color{purple}{a_{2n}a_{2n-1}-a_{2n+1}a_{2n-1}}=1 \\ a_{2n}a_{2n+1}-a_{2n-1}a_{2n+2}= 1 >0 \end{eqnarray*} Now this will rearrange to give \begin{eqnarray*} \frac{a_{2n+1}}{a_{2n+2}}>\frac{a_{2n-1}}{a_{2n}}. \end{eqnarray*} The increasing of the sequence for odd terms can be shown in a similar manner.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .