Let $a \in (0,1)$ be number with decimal representation. Where $a_k \in \{0,1,...,9\}\: for\: k \in\mathbb{N}\}$. Show that the sequence $(x_n)$ with $x_n = \sum_{k=1}^{n} a_k 10^{-k}$$ \;,\: n \in\mathbb{N}$ is a Cauchy sequence.
I've tried using the triangle inequality but it didn't work out
On
Note that the limit of $(x_n)$ is the same as the sum $\sum_{k=1} ^{\infty} a_k 10^{-k}$. This is because the sum of the series $\sum_{k=1} ^{\infty} a_k 10^{-k}$ is by definition the limit of the $n^{th}$ partial sum of that series which is $(x_n)$.
Since $a_k\leq9$ $\forall$ $k\in\mathbb{N}$, we can write
$\sum_{k=1} ^{\infty} a_k 10^{-k} \leq \sum_{k=1} ^{\infty} 9(( \frac{1}{10})^{k})$ $\forall$ $k\in\mathbb{N}$
The bigger series here converges as $\sum_{k=1} ^{\infty} ( \frac{1}{10})^{k}$ is a converging infinite geometric series. Hence by Comparison Test, the smaller series $\sum_{k=1} ^{\infty} a_k 10^{-k}$ also converges.
Hence, $(x_n)$ is a convergent sequence. It can be concluded to be Cauchy by invoking the fact that "A Sequence is Convergent if and only if it is Cauchy".
Note that $\sum_{k=N}^n a_k 10^{-k} < 1 (10)^{N-1} $ for what value of $n$