Prove the sequence is Cauchy and therefore Convergent Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $\mathbb{R}$. Let $r\in (0,1)$ and suppose that $|x_{n+1}-x_n|&lt;r^n \forall n\in \mathbb{N}$. Show that $(x_n)_{n\in\mathbb{N}}$ converges.

I know that it is sufficient to show that the sequence is Cauchy. So, this is very similar to this past question (Please help prove that $(x_n)$ is a Cauchy sequence if $|x_{n+1} - x_n| \leq Cr^n$) but I don't completely understand the answers given so I'm hoping someone can help me out.
Let $k &gt; 0$ be arbitrary. Then, using the triangle inequality once,
$$
|x_{n+k}-x_n| = |x_{n+k}-x_{n+k-1}+x_{n+k-1} - x_n|\le|x_{n+k}-x_{n+k-1}|+|x_{n+k-1} - x_n|.
$$
Applying it again on the last term gives
\begin{align*}
|x_{n+k}-x_n|
&amp;\le|x_{n+k}-x_{n+k-1}|+|x_{n+k-1} - x_n|\\
&amp;\le |x_{n+k}-x_{n+k-1}| + |x_{n+k-1} - x_{n+k-2}| + |x_{n+k-2}-x_n|.
\end{align*}
Using it over and over again, you end up with
$$
|x_{n+k}-x_n|\le\sum_{l=1}^k|x_{n+l}-x_{n+l-1}|.
$$
Now, use what you know:
\begin{align*}
|x_{n+k}-x_n|
&amp;\le\sum_{l=1}^k|x_{n+l}-x_{n+l-1}|\le\sum_{l=1}^k r^{n+l-1} = r^n\sum_{l=1}^k r^{l-1}\\
&amp;= r^n\sum_{l=0}^{k-1} r^{l}\le r^n\sum_{l=0}^\infty r^{l} = \frac{r^n}{1-r}.
\end{align*}
In the last equality we used the geometric series. As $k$ was arbitrary, we obtain $|x_m-x_n| &lt; \frac{r^n}{1-r}$ for each $m\ge n$. Therefore, $(x_n)$ is Cauchy.
Let $k &gt; 0$ be arbitrary. Then, using the triangle inequality once,
$$
|x_{n+k}-x_n| = |x_{n+k}-x_{n+k-1}+x_{n+k-1} - x_n|\le|x_{n+k}-x_{n+k-1}|+|x_{n+k-1} - x_n|.
$$
Applying it again on the last term gives
\begin{align*}
|x_{n+k}-x_n|
&amp;\le|x_{n+k}-x_{n+k-1}|+|x_{n+k-1} - x_n|\\
&amp;\le |x_{n+k}-x_{n+k-1}| + |x_{n+k-1} - x_{n+k-2}| + |x_{n+k-2}-x_n|.
\end{align*}
Using it over and over again, you end up with
$$
|x_{n+k}-x_n|\le\sum_{l=1}^k|x_{n+l}-x_{n+l-1}|.
$$
Now, use what you know:
\begin{align*}
|x_{n+k}-x_n|
&amp;\le\sum_{l=1}^k|x_{n+l}-x_{n+l-1}|\le\sum_{l=1}^k r^{n+l-1} = r^n\sum_{l=1}^k r^{l-1}\\
&amp;= r^n\sum_{l=0}^{k-1} r^{l}\le r^n\sum_{l=0}^\infty r^{l} = \frac{r^n}{1-r}.
\end{align*}
In the last equality we used the geometric series. As $k$ was arbitrary, we obtain $|x_m-x_n| &lt; \frac{r^n}{1-r}$ for each $m\ge n$. Therefore, $(x_n)$ is Cauchy.