To show that a recursively defined sequence by $x_1=\frac12$ and $x_{n+1}=\frac{x_n^3 + 2}{7}$ is Cauchy - How? 
Show that the sequence defined by $x_1$ = $\frac{1}{2}$ and $x_{n+1} = \frac{x_n^3 + 2}{7}$ for $n \in N$ satisfies the Cauchy criterion.

I don't understand how to go about this problem. Is it alright to show that the sequence is convergent, which would then imply it being Cauchy? or is there some way to directly get expressions in terms of the Cauchy criteria?
My observations:
All terms of the sequence are positive, and we can write $|x_{m+1}-x_{n+1}| = |\frac{x_m^3 - x_n^3}{7}|$
How do I show that for an arbitrarily chosen $\epsilon$, the expression is less than $\epsilon$ for all $m,n \ge$ some natural number $K$?
Could someone please help me with the solution, or possibly point me in the right direction? Thanks.
 A: First, show by induction that, with $x_1=1/2$, it follows that $0<x_n\leq 1/2$. Then, since $(x^3 - y^3)=(x^2+xy+y^2)(x-y)$, we find that
$$|x_{n+1}-x_{n}| = \frac{|x_n^3 - x_{n-1}^3|}{7}\leq\frac{\frac{1}{2^2} + \frac{1}{2^2}+\frac{1}{2^2}}{7} |x_{n}-x_{n-1}|=\frac{3}{28}|x_{n}-x_{n-1}|.$$
Now use the triangle inequality to estimate $|x_n-x_m|$. Can you take it from here?
A: You can prove by induction that $0<x_n< 1$ for all $n \in \mathbb N$.
The derivative of the function $f(x)=\frac{x^3+2}{7}$ is $f^\prime(x) = \frac{3}{7}x^2$. Hence $0\le f^\prime(x)\le  \frac{1}{2}$ for $0 \le x \le 1$.
Using the mean value theorem, you get
$$\vert f(x)-f(y) \vert \le \frac{1}{2}\vert x -y \vert$$ for $x,y \in [0,1]$.
Which leads to
$$\vert x_{n+1}-x_n \vert \le \frac{1}{2} \vert x_n - x_{n-1} \vert $$ for all $ n \in \mathbb N$.
You then conclude that the sequence is Cauchy by induction.
This is a case of a more general result when the function is a contraction mapping.
A: By induction we see that $0<x_n<1$ for all $n$. Hence $|x_{m+1}-x_m| \leq \frac  3 7 |x_m-x_{m-1}|$ (by MVT theorem for $x \to x^{3}$). This gives $|x_{m+1}-x_m| \leq (\frac 3 7)^{m-1} |x_2-x_1|$. This implies that $\sum |x_{m+1}-x_m| <\infty$. Can you take it from here?
