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.
 A: Hint:
You van use this formula:
$$x_n\leqslant\big(\frac53\big)^n$$
$$\frac{x_{n+1}}{x_n}\leqslant\frac53$$
Note: The proof of this formula is in "Discrete and Combinatorial  Mathematics, an applied introductioc" ,Ralph. P. Grimaldi
A: Note: I'm not 100% certain that this can be an acceptable answer, so I'm asking others to help me improve it, but this is what I've thought about the problem.
Writing $t_n$ as $t_n=1+\frac{1}{t_{n-1}}, n\in\Bbb{N}$ with $t_0=\infty$, we see that
$$(t_n)=\{1,2,3/2,5/3,8/5,13/8,\dots\}.$$
Like you have observed, the even-index terms are decreasing and the odd-index terms are decreasing. Let's prove this by induction. The proposition is that $t_{2n}>t_{2(n+1)}$
$$
\implies\frac{1}{t_{2n}}<\frac{1}{t_{2(n+1)}}\implies 1+\frac{1}{t_{2n}}<1+\frac{1}      {t_{2(n+1)}}\implies t_{2n+1}<t_{2(n+1)+1}
$$
This proves that the odd-index terms are increasing. Similarly, we start with the proposition $t_{2n-1}<t_{2n+1}$
$$
\implies \frac{1}{t_{2n-1}}>\frac{1}{t_{2n+1}}\implies 1+\frac{1}{t_{2n-1}}>1+\frac{1}{t_{2n+1}}\implies t_{2n}>t_{2(n+1)}
$$
which proves that the even-index terms are decreasing.
Observe that $0<t_{2n}<2$ and $1\leq t_{2n-1}<2$. Thus these are bounded and hence by the Monotone Convergence Theorem, the subsequences converge. Since these are subsequences of the given sequence $(t_n)$, we can conclude that $(t_n)$ also converges.
A: $$X_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]=\frac{a^n-b^n}{\sqrt{5}}$$
So $\lim_{n \to \infty} \frac{X_{n+1}}{X_n}=a=\frac{1|+\sqrt{5}}{2}$
