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Question Number 72671 by aliesam last updated on 31/Oct/19

Commented by kaivan.ahmadi last updated on 31/Oct/19

a) ∣x_(n+1) −x_n ∣≤c∣x_n −x_(n−1) ∣≤c^2 ∣x_(n−1) −x_(n−2) ∣≤...≤c^n ∣x_1 −x_0 ∣ b) ∣x_(n+1) −x_0 ∣=∣(x_(n+1) −x_n )+(x_n −x_(n−1) )+...+(x_1 −x_0 )∣≤ ∣x_(n+1) −x_n ∣+∣x_n −x_(n−1) ∣+...+∣x_1 −x_0 ∣≤ c^n ∣x_1 −x_0 ∣+c^(n−1) ∣x_1 −x_0 ∣+...+∣x_1 −x_0 ∣= (c^n +c^(n−1) +...+1)∣x_1 −x_0 ∣ c) x_(n+1) −x_k =(x_(n+1) −x_n )+(x_n −x_(n−1) )+...+(x_(k+1) −x_k )⇒ c is clear.

a)∣xn+1−xn∣⩽c∣xn−xn−1∣⩽c2∣xn−1−xn−2∣⩽...⩽cn∣x1−x0∣b)∣xn+1−x0∣=∣(xn+1−xn)+(xn−xn−1)+...+(x1−x0)∣⩽∣xn+1−xn∣+∣xn−xn−1∣+...+∣x1−x0∣⩽cn∣x1−x0∣+cn−1∣x1−x0∣+...+∣x1−x0∣=(cn+cn−1+...+1)∣x1−x0∣c)xn+1−xk=(xn+1−xn)+(xn−xn−1)+...+(xk+1−xk)⇒cisclear.

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