Fibonacci sequence and convergence of $t_n=\cfrac{x_{n+1}}{x_n}$ Fibonacci sequence define as: $x_1=x_2=1 , x_n=x_{n-1}+x_{n-2}$ for $n \ge 1$ prove the sequence $t_n=\cfrac{x_{n+1}}{x_n}$ is convergent.
First of all I start by calculating some of the first terms of the $t_n$ sequence:
$\begin{array}{rcc}
n:&1&2&3&4&5&6\\
t_n:&\frac11&\frac21&\frac32&\frac{5}3&\frac{8}{5}&\frac{13}{8}
\end{array}$
It seems for even values of $n$ the sequence is decreasing and for odd values of $n$ it is increasing.
To prove the first statement (where $n=2k$) , I should prove $t_{2k}- t_{2k+2}\ge 0$ :
$$\cfrac{x_{2k+1}}{x_{2k}}- \cfrac{x_{2k+3}}{x_{2k+2}} \ge0$$
$$\cfrac{x_{2k+1} \cdot x_{2k+2}- x_{2k+3} \cdot x_{2k}}{x_{2k} \cdot x_{2k+2}} \ge 0$$
$$x_{2k+1} \cdot x_{2k+2} \ge x_{2k+3} \cdot x_{2k}$$
$$x_{2k+1} \cdot (x_{2k}+x_{2k+1}) \ge  x_{2k} \cdot (x_{2k+1}+x_{2k+2})$$
$$x_{2k+1} \times x_{2k+1} \ge x_{2k} \times x_{2k+2}$$
Here I don't khow how to proceed.

$$X_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]=\frac{a^n-b^n}{\sqrt{5}}$$
So $\lim_{n \to \infty} \frac{X_{n+1}}{X_n}=a=\frac{1|+\sqrt{5}}{2}$
