26A03-IfDxiXi112iThenXiIsACauchySequence.tex

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\pmtitle{if $d(x_i, x_{i+1})<1/2^i$ then $x_i$ is a Cauchy sequence}
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\begin{document}
\begin{lemma}
Suppose $x_1, x_2, \ldots$, is a sequence in a metric space. 
If for some $N\ge 1$, we have $d(a_i,a_{i+1})<1/2^i$ for all $i\ge N$, 
then $\{x_i\}$ is a Cauchy sequence. 
\end{lemma}

\begin{proof} 
Let us denote by $d$ the metric function. 
If $\varepsilon>0$, then for some $N\in \sN$ we have $1/2^N<\varepsilon$.
Thus, if $N<m<n$ we have
\begin{eqnarray*}
d(x_m, x_n) &\le& d(x_m, x_{m+1}) + \cdots + d(x_{n-1}, x_n) \\
   &=& \left(\frac{1}{2}\right)^m + \cdots + \left(\frac{1}{2}\right)^{n-1} \\
   &=&  \left(\frac{1}{2}\right)^{m-1} \ \sum_{i=1}^{n-m} \left(\frac{1}{2}\right)^i \\
   &=&  \left(\frac{1}{2}\right)^{m-1} \ \frac{1-\left(\frac{1}{2}\right)^{n-m}}{1-\frac{1}{2}} \\
   &<&  \left(\frac{1}{2}\right)^{m} \\
   &<&  \left(\frac{1}{2}\right)^{N} \\
   &<&  \varepsilon,
\end{eqnarray*}
where we have used the triangle inequality and the 
\PMlinkname{geometric sum formula}{GeometricSeries}.
\end{proof}
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