About a proof that the sequence $ x_{n+1}=\frac{x_n+1}{x_n+2}$ is Cauchy. I have an exercise problem, which explicitly sais:
Show that the sequence $\left(x_n\right)_{n=0}^{\infty}$ which is defined recursively by
$$
x_{n+1}=\frac{x_n+1}{x_n+2},\space x_0=1
$$
is Cauchy and determine its limit.
I proceeded as follows:
By induction, we can show that for all $n\in\mathbb{N_0}$:
$$
0<x_{n+1}<x_n
$$
Thus:
$$
x_n-x_{n+1}=\frac{\left(x_{n-1}-x_n\right)}{\left(x_n+2\right)\left(x_{n-1}+2\right)}<\frac{1}{4}(x_{n-1}-x_n)<...<\frac{1}{4^n}\left(x_0-x_1\right)=\frac{1}{3\cdot4^n}
$$
Therefore, for $r\in\mathbb N$:
$$
x_n-x_{n+r}=\sum_{k=0}^{r-1}x_{n+k}-x_{n+k+1}<\sum_{k=0}^{r-1}\frac{1}{3\cdot 4^{n+k}}<\sum_{k=0}^{\infty}\frac{1}{3\cdot 4^{n+k}}=\frac{1}{9\cdot4^{n-1}}
$$
From which we can easily deduce that the sequence is Cauchy. Then to solve for the limit isn't difficult.
What bothers me about this proof, is that for the existence of the limit, we only need $0<x_{n+1}<x_n$. Is there a faster way to show that it is Cauchy? Or is the intuition behind the exercise to practice the definitions, regardless of wether showing that a sequence is Cauchy is the fastest way to proceed?
Thanks in advance.
 A: You’re quite right that you already have convergence from the fact that it’s monotone and bounded. Like Paul Sinclair, I suspect that the point of the exercise is to give you practice in showing directly that a sequence is Cauchy. I consider it a bad exercise, since it asks you to do something that is completely unnatural in this setting. Just for fun, here’s a way to prove directly that it’s convergent and simultaneously derive the limit, given a little background knowledge.
Clearly each $x_n$ is rational, so suppose that $x_n=\frac{a_n}{b_n}$, where $a_n$ and $b_n$ are integers. We can set $a_0=b_0=1$, and the recurrence becomes
$$\frac{a_{n+1}}{b_{n+1}}=\frac{\frac{a_n}{b_n}+1}{\frac{a_n}{b_n}+2}=\frac{a_n+b_n}{a_n+2b_n}\;,$$
giving us recurrences
$$a_{n+1}=a_n+b_n\qquad\text{and}\qquad b_{n+1}=a_n+2b_n\;.\tag{1}$$
Write out the first few terms:
$$\begin{array}{rcc}
n:&0&1&2&3&4&5&6\\ \hline
a_n:&1&2&5&13&34&89&233\\
b_n:&1&3&8&21&55&144&377
\end{array}$$
Those numbers are very familiar: they’re the Fibonacci numbers $F_1$ through $F_{14}$. This suggests that 
$$a_n=F_{2n+1}\qquad\text{and}\qquad b_n=F_{2n+2}\;.$$
This is very easy to confirm by induction from the recurrences in $(1)$, and it follows that 
$$x_n=\frac{F_{2n+1}}{F_{2n+2}}$$
for each $n\in\Bbb N$. It’s well known that $\lim\limits_{n\to\infty}\frac{F_n}{F_{n+1}}=\frac1\varphi=\varphi-1=\frac12(-1+\sqrt5)\approx0.618$.
