Below is my attempt at doing this, would love to get some feedback on it and find out if there is anything I can improve!
For two terms in succession (for each integer $n\geq 2$):
$ \displaystyle | x_{n+1} - x_{n}| = \frac { | x_{n} - x_{n-1} | }{ (2+x_n) (2+x_{n-1} ) } < \frac 14 | x_n - x_{n-1} | < \frac 1{4^2} | x_{n-1} - x_{n-2} | < \ldots < \frac 1{4^{n-1}} | x_{n-(n-2)} - x_{n-(n-1)} | = \frac 1{4^ {n-1} } | x_2 - x_1| = \frac {c_1}{4^{n-1} }. $
Here $c_1 = |x_2 - x_1| = | \frac 25 - \frac 12| > 0$.
Given $\epsilon > 0$, we want to find an integer $N \geq 1$ such that for each integer $n,m \geq N$ we have $|x_n - x_m| < \epsilon$.
Without loss of generality suppose $m > n \geq N$, where $N$ will soon be chosen judiciously.
The distance from $x_m$ to $x_n$ is less than the sum of the distances of successive terms starting at $n$ up to $m$. Namely:
$ |x_m - x_n| \leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \ldots + | x_{n+2} - x_{n+1} | + | x_{n+1} - x_n | . $
We can control each term since we know distances of successive terms:
$ \displaystyle |x_m - x_n | \leq \frac {c_1}{ 4^{m-2} } + \frac {c_1}{ 4^{m-3} } + \ldots + \frac {c_1}{ 4^n } + \frac {c_1}{ 4^{n-1} } = \frac {c_1}{ 4^{n-1} } \cdot \left( \frac 1{4^{m-n-1}} + \frac 1{4^{m-n-2}} + \ldots + \frac 1{4} + 1 \right) $
The latter factor is a finite geometric sum with $0 < x < 1$ (in our case $x = 1/4$) and can be bounded above by a constant:
$ \displaystyle |1 + x+ x^2 + \ldots + x^n| = \frac {|1 - x^{n+1}|}{1-x} \leq \frac 2{1-x} \equiv c_2 $
Thus:
$ \displaystyle |x_m - x_n | \leq \frac {c_1}{ 4^{n-1} } \cdot \left( \frac 1{4^{m-n-1}} + \frac 1{4^{m-n-2}} + \ldots + \frac 1{4} + 1 \right) < \frac {c_1 \cdot c_2}{4^{n-1}} < \frac {c_1 \cdot c_2}{4^{N-1}} $
This upper bound depends on $N$ and by choosing it large enough we can make it smaller than the given $\epsilon$ (by Archimedean property).
Hence an integer $N$ exists for which $|x_m - x_n| < \epsilon$, which establishes the fact that the sequene is Cauchy. Done. $\blacksquare$
Any thoughts?
