Let $k$ be a positive real number with $k < 1$, and let $f$ be a $k$-contraction. Let $c\in \mathbb{R}$ be given, and define a sequence $(x_n)n≥1$ by $x_1 = c$ and $x_{n+1} = f(x_n)$. Set $C = |x_1-x_2 |$. By induction on $n$ prove that $|x_n-x_{n+1}|\leq k^{n-1}C$ for every $n\geq 1$. Then use the comparison test to show that the series $\sum_{i=1}^∞(x_{i+1}-x_i)$ converges absolutely.
can someone give me a hint or a guide...
Base step: $n = 1$ $$ |x_1-x_2| = C \leq k^{1-1}C .$$
Inductive step: $$ |x_{n+1} - x_{n+2}| = | f(x_n) - f(x_{n+1})| \leq k | x_n - x_{n+1}| \leq k k^{n-1} C |x_1-x_2| = Ck^{(n+1) -1} |x_1-x_2|. $$
The first inequality holds because $f$ is a $k$-contraction, the second because of the inductive hypothesis.
Then $$ \sum_{i=1}^{\infty}|x_{i+1} - x_i| \leq |x_2-x_1| \sum_{i=1}^{\infty} k^{n-1} = |x_2-x_1| \sum_{i=0}^{\infty} k^{n} = |x_2-x_1| \frac{1}{1-k} .$$ The last sum is a geometric series which converges since $ k < 1$. Hence the series converges absolutely.
