A sequence whose consecutive elements get closer faster then a geometric progression is a Cauchy sequence Let $K \in (0,1)$, and $d(x_{n+1}, x_n) \leq K^n l$ where $l = d(x_0, x_1)$. That is, let $\{x_n\}$ be a sequence whose consecutive elements don't get apart faster than a geometric progression that converges (that's why $K \in (0,1)$.
How can I prove that such sequences are Cauchy sequences? I know it intuitively, as the elements keep getting closer, which is the definition of a Cauchy sequence, but I'm having a hard time to prove it formally...
To show that it is indeed Cauchy, we have for $n&gt;m$:
$$\begin{align}d(x_m, x_n) &amp;\le d(x_m, x_{m+1}) + d(x_{m+1}, x_{m+2}) + \cdots + d(x_{n-1},x_n)
\\~\\&amp;\le K^ml+K^{m+1}l+\cdots + K^{n-1}l
\\~\\&amp;&lt; \sum_{k=m}^\infty K^kl
\\~\\&amp;= \frac l{1-K}K^m
\end{align}$$
for sufficiently large $m$, $d(x_m, x_n)$ can be as small as we want.
Hint: A geometric series $\sum_{n\ge0}k^n$ converges for $k\in(0,1)$, hence is Cauchy.