Proof verification that $\{x_n\} = 0,\underbrace{77\dots 7}_{\text{n times}}$ is a Cauchy sequence.

Question

Given a sequence $\{x_n\}$: $$ x_n = 0,\underbrace{77\dots 7}_{\text n\ times} $$ Prove that $\{x_n\}$ is a Cauchy sequence.

Recall the definition of a fundamental sequence: $$ x_n\ \text{is fundamental} \ \iff \forall \epsilon>0 \exists N\in \Bbb N: \forall n, m >N\implies |x_n - x_m| < \epsilon $$

Rewrite $x_n$: $$ x_n = {7\over 10^1} + {7\over 10^2} + \cdots + {7\over 10^n} = \sum_{k=1}^n \frac{7}{10^k} $$

By geometric series sum: $$ x_n = \sum_{k=1}^n \frac{7}{10^k} = \frac{7}{9}\left(1 - {1\over 10^n}\right) \\ x_m = \sum_{k=1}^m \frac{7}{10^k} = \frac{7}{9}\left(1 - {1\over 10^m}\right) \\ $$

Suppose $m > n$: $$ \begin{align} |x_n - x_m| &= |x_m - x_n| = \\ &= \left|\frac{7}{9}\left(1 - {1\over 10^m}\right) - \frac{7}{9}\left(1 - {1\over 10^n}\right)\right| = \\ &= \left|\frac{7}{9}\left(1 - {1\over 10^m} - 1 + {1\over 10^n}\right)\right| = \\ &= \left|\frac{7}{9}\left({1\over 10^n} - {1\over 10^m}\right)\right| \le \left|\frac{7}{9}{1\over 10^n}\right| \le \frac{7}{9\cdot 10^N} < \epsilon \end{align} $$

This shows we've found $N$ which depends on $\epsilon$ and satisfies the definition of a Cauchy sequence.

This is the first time I'm dealing with proving a sequence is fundamental, could someone please verify whether my proof is valid?

Answer 1

Yes, just write it in the forward direction.

Suppose $N > \log_{10}\left(\frac{7}{9 \epsilon}\right)$, then for any $m,n \in \mathbb{Z}$ such that $m> n > N$, then we have $|x_n-x_m|< \epsilon$.

Answer 2

Since my proof in OP is correct I would like to add a generalization also. Consider the following $x_n$: $$ x_n = a + aq + aq^2 + \cdots + aq^{n-1} $$

Using geometric series sum for $|q| < 1$: $$ x_n = \frac{a(1-q^n)}{1-q} $$

Since $|q| < 1$ we may rewrite it as: $$ q = \frac{1}{1+r},\ r \in \Bbb R_{>0} $$

Then for $m > n$: $$\begin{align} |x_m - x_n| &= \left|\frac{a}{1-q} \left(q^n - q^m\right)\right|\\ &= \left|\frac{a}{1-q} \left(\frac{1}{(1+r)^n} - \frac{1}{(1+r)^m}\right)\right| \\&\le \left|\frac{a}{1-q} \left(\frac{1}{(1+r)^n}\right)\right| \\ & \le \frac{a}{1-q} \left(\frac{1}{(1+r)^N}\right) < \epsilon\end{align} $$

Which shows any sequence of such kind is Cauchy. Or with direct statement: $$ \frac{1-q}{a} \left((1+r)^N\right) > {1\over \epsilon}\\ (1+r)^N > \frac{a}{(1-q)\epsilon} \\ \exists N >\log_{1+r}\frac{a}{(1-q)\epsilon}, \forall m>n>N \implies |x_m - x_n| < \epsilon $$