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This question already has an answer here:

$(A,d)$ a metric space.
Let $(a_n)_{n\geq 0}$ be a sequence such that there exists a $k<1$ and for all $n\geq 1$ we get $d(a_{n+1},a_n)\leq kd(a_n,a_{n-1})$.

  1. How do I prove that $(a_n)_{n\geq 0}$ is a Cauchy sequence?

What I know:
To prove Cauchy we need to find for all $\epsilon$ an $N\geq 0$ such that for all $m,n\geq N$ we have $d(a_m,a_n)<\epsilon$. Which $N$ do I take?

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marked as duplicate by Henno Brandsma, Hagen von Eitzen, SchrodingersCat, Alex M., Shaun Feb 20 '16 at 18:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @No, it's not. That's another question $\endgroup$ – user316173 Feb 20 '16 at 12:55
  • $\begingroup$ @gniourf_gniourf But cauchy means there exists an $N\geq0$ such that $\forall m,n\geq N:d(a_m,a_n)<\epsilon$, how does that follow? $\endgroup$ – user316173 Feb 20 '16 at 12:58
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    $\begingroup$ $a_n=\sum_{j=1}^{n}k^j$ is Cauchy and $\sum_{m+1}^n k^j=a_n-a_m$ $\endgroup$ – Peter Melech Feb 20 '16 at 13:03
  • $\begingroup$ @TheGovernator not before the edit!! The upvoted answer still refers to the original formulation. This question is just reintroducing it again. $\endgroup$ – Henno Brandsma Feb 20 '16 at 13:13
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Assume that $d(a_1,a_0)\neq0$ (otherwise the proof is obvious as the sequence $(a_n)_{n\in\mathbb{N}}$ is constant).

First show by induction that: $$\forall n\in\mathbb{N},\ d(a_{n+1},a_n)\leq k^n d(a_1,a_0).$$

Then, let $m,n\in\mathbb{N}$ with $m<n$, and observe that, by the Triangle Inequality: $$d(a_n,a_m)\leq\sum_{j=m}^{n-1} d(a_{j+1},a_j)\leq d(a_1,a_0)\sum_{j=m}^{n-1} k^j\leq d(a_1,a_0)\frac{k^m}{1-k}.$$

Now, since $\lvert k\rvert<1$, we know that $$\lim_{m\to+\infty}\frac{k^m}{1-k}=0.$$

Let $\varepsilon>0$. From the previous limit, we conclude that there exists $N\geq0$ such that $$\forall m\geq N,\ 0\leq\frac{k^m}{1-k}<\frac{\varepsilon}{d(a_1,a_0)}.$$ Hence, for all $m,n\in\mathbb{N}$ such that $N\leq m<n$ one has $$d(a_n,a_m)\leq d(a_1,a_0)\frac{k^m}{1-k}\leq d(a_1,a_0)\frac{\varepsilon}{d(a_1,a_0)}=\varepsilon.$$

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