if then $x_i$ is a Cauchy sequence

Lemma 1.

Suppose $x_{\mathrm{1}}\mathrm{,}{x}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }$, is a sequence in a metric space.If for some $N\mathrm{\ge }\mathrm{1}$, we have for all $i\mathrm{\ge }N$,then $\mathrm{\{}{x}_i\mathrm{\}}$ is a Cauchy sequence.

Proof.

Let us denote by $d$ the metric function.If $\epsilon >0$, then for some $N\in \mathbb{N}$ we have .Thus, if we have

$d(x_m,x_n)$$\le$$d(x_m,x_{m+1})+\mathrm{\cdots }+d(x_{n-1},x_n)$

$=$${\left({\displaystyle \frac{1}{2}}\right)}^m+\mathrm{\cdots }+{\left({\displaystyle \frac{1}{2}}\right)}^{n-1}$

$=$${\left({\displaystyle \frac{1}{2}}\right)}^{m-1}{\displaystyle \sum _{i=1}^{n-m}}{\left({\displaystyle \frac{1}{2}}\right)}^i$

$=$${\left({\displaystyle \frac{1}{2}}\right)}^{m-1}{\displaystyle \frac{1-{\left(\frac{1}{2}\right)}^{n-m}}{1-\frac{1}{2}}}$

${\left({\displaystyle \frac{1}{2}}\right)}^m$

${\left({\displaystyle \frac{1}{2}}\right)}^N$

$\epsilon ,$

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