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It is given that $(X,d)$ is a metric space. Suppose that $(x_n)_{n∈\Bbb N}$ is a sequence in X with the property that $\sum _{n=1} ^ \infty d(x_{n+1},x_n) < \infty$. I am asked to show that the sequence is convergent. I decided to prove this by showing that this sequence is a Cauchy sequence in X. In other words to show that for every $\epsilon >0$ there exists $N \in \Bbb N$ such that for all n,m $\ge$N $d(x_n,x_m) < \epsilon $.

However I don't know how to start and how to use the given property.

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$d(x_{n+m},x_n)\leq d(x_{n+1},x_n)+...+d(x_{n+m},x_{n+m-1})$, since $\sum_nd(x_{n+1},x_n)$ finite, for every $c>0$, there exists $n_0$ such that $n>n_0$ implies that $d(x_{n+m},x_n)\leq d(x_{n+1},x_n)+...+d(x_{n+m},x_{n+m-1})<c$, thus $x_n$ is a Cauchy sequence.

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This is quite simplly done, let $$c=\sum_{i=1}^\infty d(x_n,x_n+1)$$ From this do we know that $$\lim_{n\to\infty}c-\sum_{i=1}^n d(x_n,x_n+1)=0$$ which means that for any $\epsilon$ we can find an $N$ such that $n,m>N$ we have that $d(x_n,x_m)<\epsilon$

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  • $\begingroup$ do you mean $x_{n+1} $ ? @Zelos $\endgroup$ – user189013 Nov 2 '15 at 16:45
  • $\begingroup$ I mean $x_m$ as that is what is required for it to be a cauchy sequence. $\endgroup$ – Zelos Malum Nov 2 '15 at 16:46

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