Let $a \in (0, 1)$ be a number with decimal representation:
$0.a_1 a_2 a_3 \ldots,$
where $a_k \in \{0,1,\ldots,9\}$ for $k \in \mathbb{N}$. Show that the sequence $(x_n)$ with:
$x_n = \sum_{k=1}^{n} a_k \cdot 10^{-k}$
is a Cauchy sequence.
That what I did:
For any $\varepsilon > 0$, we first conclude that there exists some $N \in \mathbb{N}$ such that $0 < \varepsilon$. Now let $m, n \geq N$. Without loss of generality, we assume that $m > n$, say $m = n + k$ for some $k \in \mathbb{N}$. Then,
$x_m - x_n = \sum_{k=1}^{m} a_k \cdot 10^{-k} - \sum_{k=1}^{n} a_k \cdot 10^{-k} = \sum_{k= n + 1}^{m} a_k \cdot 10^{-k} < \varepsilon$.
But I don't have any ideas what to do next.
I'd replace the first sentence in your proof by: "For every $\epsilon > 0$ there is some $N_\epsilon$ such that $10^{-N_\epsilon}<\epsilon$, and therefore $10^{-n}<\epsilon$ for all $n>N_\epsilon$.
With this, we see that for any choice of $m>n > N_\epsilon$, we get $$|x_n - x_m| = \sum_{j=n+1}^ma_j10^{-j} < 10^{-n} < 10^{-N_\epsilon} < \epsilon.$$ This completes the proof.