Fibonacci sequence and convergence of $t_n=\cfrac{x_{n+1}}{x_n}$

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

Did anybody recall Cassini's identitiy? $$x_n^2=x_{n+1}x_{n-1}+(-1)^{n+1}$$ For odd $n$, $$x_{2k+1}^2=x_{2k}x_{2k+2}+1$$