a sequence {xn} of real numbers is contractive if ∃ a constant C > 0, 0 < C < 1, such that |xn+1 − xn| ≤ C|xn − xn−1| for all n ∈ N.

Question

We say that a sequence {xn} of real numbers is contractive if ∃ a constant C > 0, 0 < C < 1, such that |xn+1 − xn| ≤ C|xn − xn−1|

for all n ∈ N. Answer the following: Show that every contractive sequence is convergent.

Answer 1

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>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.