Prove that $a_1 ≤ a_3 ≤ \cdots ≤ a_{2n+1} ≤ a_{2n} ≤ a_{2n-2} ≤ \cdots ≤ a_2$ where $a_i=f_{i+1}/f_i$ and $\{f_i\}$ is the Fibonacci sequence Let $\displaystyle a_i := \frac{f_{i+1} }{f_i}$, where $\{f_i\}$ is the Fibonacci sequence. Prove by induction that $\forall n ≥ 1 $,
$$a_1 ≤ a_3 ≤ \cdots ≤ a_{2n+1} ≤ a_{2n} ≤ a_{2n-2} ≤ \cdots ≤ a_2$$
From what I understand, since Fibonacci is $f_{n-2} + f_{n-1}$ then $f_{i+1}$ should be $f_{n-1} + f_n$. 
From there, I'm completely lost on where to go to prove this by induction.
 A: The basis cases are easy to check (put the numbers in!) For the inductive step:
By the definition of the Fibonacci numbers, for any integer $k>0$ and $m \geq 0$ we have
$$ a_{k+1}-a_{k} = \frac{f_{k+2}}{f_{k+1}} - \frac{f_{k+1}}{f_{k}} \\
= \frac{f_{k+1}+f_{k}}{f_{k+1}} - \frac{f_{k}+f_{k-1}}{f_k} \\
= \frac{f_{k}}{f_{k+1}} - \frac{f_{k-1}}{f_{k}} \\
= \frac{1}{a_{k}} - \frac{1}{a_{k-1}} \\
= \frac{1}{a_{k}a_{k-1}} (a_{k-1}-a_{k}) . $$
In particular, since every Fibonacci number is positive, $a_{k+1}-a_k$ has the same sign as $a_{k-1}-a_{k}$. In particular, by induction, the following signs are the same:
$$ a_2 - a_1 , a_2 - a_3 , a_4 - a_3 , a_4 - a_5 , \dotsc , a_{2n-2}-a_{2n-1} , a_{2n} - a_{2n-1} , a_{2n} - a_{2n+1} , a_{2n+2} - a_{2n+1} , \dotsc $$
It is easy to check that $a_2-a_1$ is positive, and so by induction,
$$ a_2 > a_1 , a_2 > a_3 , a_4 > a_3 , a_4 > a_5 , \dotsc , a_{2n-2} > a_{2n-1} , a_{2n} > a_{2n-1} , a_{2n} > a_{2n+1} , \dotsc . $$
A similar calculation shows that $a_{k+2}-a_k$ has the same sign as $a_{k-2}-a_k$, which on noting that $a_4<a_2$ and $a_1<a_3$, implies that
$$ a_1 < a_3 < a_5 < \dotsb a_{2n-1} < a_{2n+1} < \dotsb $$
and
$$ \dotsb < a_{2n} < a_{2n-2} < \dotsb < a_6 < a_4 < a_2 . $$
But $ a_{2n+1} < a_{2n}$ by the above, and the result follows.
A: It follows from Cassini's identity:
$$
f_{n-1}f_{n+1} - f_n^2 = (-1)^n
$$
Apply this to
$$
\frac{a_{i+2}}{a_i}
=
\frac{f_{i+2}}{f_{i+1}}
\frac{f_i}{f_{i+1}}
$$
