$\left|x_{n+1} - x_n\right| < C\left|x_n - x_{n-1}\right|$ prove this is Cauchy Suppose that for the sequence $\{x_n\}$ there exists $0 < C < 1$ such that 
$$\left|x_{n+1} - x_n\right| < C\cdot \left|x_n - x_{n-1}\right|.$$  Prove that ${x_n}$ is Cauchy. 
 A: Use condition several times
$$|x_{n+1}-x_n|\le C |x_n-x_{n-1}|\le C^2|x_{n-1}-x_{n-2}|\le...\le C^n|x_1-x_0|$$
and for $\;p>0\;$
$$|x_{n+p}-x_n|=|x_{n+p}-x_{n+p-1}+x_{n+p-1}-x_{n+p-2}+...+x_{n+1}-x_n|\le$$
$$\le\sum_{k=0}^{p-1}|x_{n+p-k}-x_{n+p-1-k}|\le\sum_{k=0}^{p-1}|x_1-x_0|C^{n+p}$$
and last sum is converging since it is geometric and $\;0<C<1\;$ 
A: Sketch of proof:


*

*Prove that $\lvert x_{n+k}-x_{n+k-1}\rvert\le C^k\lvert x_n-x_{n-1}\rvert$

*Use the above to prove that $\lvert x_{n+m}-x_n\rvert\le \frac{1-C^{m}}{1-C}\lvert x_{n+1}-x_n\rvert\le\frac{1}{1-C}\lvert x_{n+1}-x_n\rvert$

*Use these inequalities (in particular, the first one with $n=1$) to prove the sequence is Cauchy.
A: We see that $ x_{n+1}- x_n \le c|x_n-x_{n-1}|
                          \le c^2|x_{n-1}-x_{n-2}|
                           \le ...
                           \le c^{n-1}|x_2-x_1|$
Let$ m\gt n$
Then$ |u_m-u_n|\le|u_m -u_{m-1}|+...+|u_{n+1}-u_n|
              \le|u_2-u_1|\{c^{m-3}+c^{m-4}+...+c^{n-2}\}
              \le|u_2-u_1|c^{n-2}.\frac{1-c^{m-n-1}}{1-c}
              \le\frac{c^{n-2}}{1-c}|u_2-u_1|$
 Let$\epsilon >0$, 
Since$0 \lt c\lt1$,the sequence $\{c_{n-2}\}$ is a convergent sequence.
Therefore there exists a natural number P such that 
$$\frac{c^{n-1}}{1-c}|u_2-u_1|\lt\epsilon$$ for all m,n$\gt P$
 It follows that $|u_m-u_n|\lt\epsilon$ for all m,n $\gt P$ and this proves that,$\{x_n\}$ is a Cauchy sequence and it's convergent.
