Examine the convergence of the following sequence If $x_1=6$ and $x_{n+1}=5 -\frac{6}{x_n} ; n\geq1$ then examine the convergence of this sequence.
I tried AM-GM but it leads to $ \frac{x_{n+1}}{x_n}\leq\frac{25}{24} $ from where I could not proceed.
The solution uses mathematical induction but is there any other way? Please help.
 A: Let $$f \; : \; x\mapsto 5-\frac 6x$$
$$f'(x)=\frac{6}{x^2}>0$$
$f$ is increasing from $(3,6)$ to $(3,6) $
$$x_2=4<x_1 \implies (x_n) \text{ is decreasing}$$
on the other hand
$$f(x)=x \implies x^2-5x+6=0$$
$$\implies x=2 \text{ or } x=3$$
but
$$x_1=4>3 \implies x_n>3$$
we conclude that $(x_n)$ converges to $3$.
A: Let $x_{n+1} = \frac{5x_n - 6}{x_n}$.
Then $\frac{x_{n+1}}{x_n} = \frac{5x_n-6}{x_n^2} $ call it $(1)$.
We'll first prove $x_n > 3$
Clearly $x_1 > 3$. Then assuming $x_n > 3$ we have: $x_{n+1} = 5 - \frac{6}{x_n} > 5 - \frac{6}{3} = 3$.
As we know that, going back to $(1)$, we'll prove : $\frac{5x_n - 6}{x_n^2} < 1$.
It is equivalent to $x_n^2 - 5x_n + 6 > 0 $, so $(x_n-3)(x_n-2) > 0 $, which holds because $x_n > 3$.
That means $\frac{x_{n+1}}{x_n} < 1$, so $(x_n)_{n \in \mathbb N}$ is decreasing and bounded, hence have limit $g$.
By recursive formula, we must have $g = \frac{5g-6}{g}$ which gives us $g \in \{2,3\}$. But since $x_n > 3$ for any $n \in \mathbb N$, then $g=3$
A: Write some elements of your sequence:
$$x_1=6\,,\;\;x_2=5-\frac66=4\,,\;\;x_3=5-\frac64=\frac72\,,\;\;x_4=5-\frac6{7/2}=\frac{23}7\;,\ldots$$
What about an "educated" guess? Say, $\;x_n\ge3\;\;\forall\,n\in\Bbb N\;$, and use induction to prove:
$$x_{n+1}=5-\frac6{x_n}\ge3\iff\frac6{x_n}\le2\iff x_n\ge3\;\;\color{green}\checkmark...\text{under the assumption that}\;\;x_n>0$$
for all $\;n\in\Bbb N\;$ (try to prove this yourself.
Then, it "seems" to be your sequence is monotone ascending and bounded above by $\;5\;$ ....Can you continue?
