Proof the sequence $(x_n)=\frac{n}{n+1}$ is contractive

Question

After proving $(x_n)=(\frac{n}{n+1}:n\in \Bbb N^+)$ is convergent, a bounded, monotone, and Cauchy sequence. Next I have to prove that this sequence is contractive. Though I have no idea if it is true or false right now. I read some examples on how to do it, most of it are about defining $x_{n+1}$ first, but I don't know how to do that for this sequence.

So I don't even know how to begin proving this.

Answer 1

Note that $$\vert x_n-x_m\vert=\dfrac{\vert n-m \vert}{(m+1)(n+1)}\le\dfrac{\vert n-m\vert}{4}.$$ Becouse the minimum value of $(m+1)(n+1)$ achieve at $n=m=1.$

Answer 2

$$|x_{n+1}-x_n| = \left|\frac {n+1}{n+2}-\frac{n}{n+1}\right| = \left|\frac 1{(n+1)(n+2)}\right|$$ hence $$\frac {|x_{n+2}-x_{n+1}|}{|x_{n+1}-x_{n}|} = \frac {(n+1)(n+2)}{(n+2)(n+3)} = \frac {n+1}{n+3} = 1-\frac 2{n+3}$$

Now, does there exist any $0 < C < 1$ such that the above is bounded above by $C$?