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Let $(X, d)$ be a complete metric space, $r∈ (0,1)$ and $\{x_n\}$ be a sequence in $X$ such that

$$d(x_{n+2}, x_{n+1})≤ rd(x_{n+1}, x_n),$$ for every $n∈ℕ$. Show that $\{x_n\}$ is a convergent sequence.

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closed as off-topic by Did, wythagoras, Zachary Selk, anomaly, Mike Pierce Jun 12 '15 at 18:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, wythagoras, Zachary Selk, anomaly, Mike Pierce
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If $x_1=x_2$ we are done. Otherwise, for every $\epsilon>0$, there is a positive integer $N$ such that $r^{N-1}\leq\epsilon(1-r)\frac{1}{d(x_2,x_1)}$. Then for any $m>n\geq N$,


$\leq d(x_m,x_{m-1})+...+d(x_{n+1},x_n)$

$\leq d(x_2,x_1)(r^{m-2}+...+r^{n-1})$

$\leq d(x_2,x_1)\frac{r^{n-1}}{1-r}$

$\leq d(x_2,x_1)\frac{r^{N-1}}{1-r}$


This shows that the sequence is Cauchy so that it converges since $X$ is complete.

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HINT: Show that the sequence is Cauchy. What can you say about $\sum_{k\ge 0}d(x_{n+k},x_{n+k+1})$? You’ll be using the triangle inequality and looking at the sum of a geometric series.

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