Example
Suppose that \(\{x_n\}\) is a sequence such that
$$
|x_{n+1} - x_n| \leq \frac{1}{2^n}
$$
Then \(\{x_n\}\) is Cauchy and hence convergent.
Proof
Let \(\epsilon > 0\) be given. Choose \(N\) by the Archimedean principle such that
$$
\begin{align*}
\frac{1}{2^{N-1}} < \epsilon
\end{align*}
$$
Then whenever \(m,n > N+1\) we have
$$
\begin{align*}
|x_m - x_n| &\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \ldots + |x_{n+1} - x_n| \\
&\leq \frac{1}{2^{m-1}} + \ldots + \frac{1}{2^{n}} \\
&= \frac{1}{2^n} \left(1+ \frac{1}{2} + \ldots + \frac{1}{2^{m-n-1}} \right) \\
&= \frac{1}{2^n} \left( \frac{1 - \left(\frac{1}{2}\right)^{m-n}}{1 - \frac{1}{2}} \right) \\
&\leq \frac{1}{2^{n-1}} \leq \frac{1}{2^N} < \epsilon. \ \blacksquare
\end{align*}
$$
References
- Lecture Notes by Professor Chun Kit Lai