# if $d(x_{i},x_{i+1})<1/2^{i}$ then $x_{i}$ is a Cauchy sequence

###### Lemma 1.

Suppose $x_{1},x_{2},\ldots$, is a sequence in a metric space. If for some $N\geq 1$, we have $d(a_{i},a_{i+1})<1/2^{i}$ for all $i\geq N$, then $\{x_{i}\}$ is a Cauchy sequence   .

###### Proof.

Let us denote by $d$ the metric function. If $\varepsilon>0$, then for some $N\in\mathbb{N}$ we have $1/2^{N}<\varepsilon$. Thus, if $N we have

 $\displaystyle d(x_{m},x_{n})$ $\displaystyle\leq$ $\displaystyle d(x_{m},x_{m+1})+\cdots+d(x_{n-1},x_{n})$ $\displaystyle=$ $\displaystyle\left(\frac{1}{2}\right)^{m}+\cdots+\left(\frac{1}{2}\right)^{n-1}$ $\displaystyle=$ $\displaystyle\left(\frac{1}{2}\right)^{m-1}\ \sum_{i=1}^{n-m}\left(\frac{1}{2}% \right)^{i}$ $\displaystyle=$ $\displaystyle\left(\frac{1}{2}\right)^{m-1}\ \frac{1-\left(\frac{1}{2}\right)^% {n-m}}{1-\frac{1}{2}}$ $\displaystyle<$ $\displaystyle\left(\frac{1}{2}\right)^{m}$ $\displaystyle<$ $\displaystyle\left(\frac{1}{2}\right)^{N}$ $\displaystyle<$ $\displaystyle\varepsilon,$

where we have used the triangle inequality    and the geometric sum formula (http://planetmath.org/GeometricSeries). ∎

Title if $d(x_{i},x_{i+1})<1/2^{i}$ then $x_{i}$ is a Cauchy sequence IfDxiXi112iThenXiIsACauchySequence 2013-03-22 14:37:31 2013-03-22 14:37:31 matte (1858) matte (1858) 5 matte (1858) Result msc 26A03 msc 54E35