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?
Best Answer
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 $|x_n-x_m|< \epsilon$.