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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 don't need a name
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  • The first sentence of your proof seems redundant. The key question you need to ask is: Given $\epsilon > 0$, what $N$ do you choose such that $|x_n - x_m|<\epsilon$ for all $m,n> N$? – DominikS Oct 30 '23 at 14:17

1 Answers1

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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.

DominikS
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