Section 5.6 Cauchy sequences
The idea here is the following. The sequence \((a_n)\) converges if and only if its terms are getting arbitrarily close to a limit. That happens if and only if its terms are getting arbitrarily close te each other. That's the notion of a Cauchy sequence.
Definition 5.6.1.
Let \((a_{n})\) be a sequence of real numbers. We say that \((a_n)\) is Cauchy if and only if for every \(\epsilon\gt 0 \text{,}\) there exists a natural number \(N\) such that for every pair of natural numbers \(m,n \gt N\text{,}\) one has \(|a_m-a_n| \lt \epsilon\text{.}\)
Theorem 5.6.2. Cauchy criterion of convergence.
Let \((a_n)\) be a sequence of real numbers. Then \((a_n)\) is Cauchy if and only if it converges.
Proof.
Assume first that \((a_n)\) is Cauchy. By Bolzano-Weierstrass, there exists a convergent subsequence \((a_{n_k})\text{;}\) let \(L\) denote its limit. We now claim that \(a_n \to L\) as \(n\to \infty\text{.}\)
To prove this, let \(\epsilon \gt 0\text{.}\) The Cauchy property implies that there exists a natural number \(N_0 \in \NN\) such that for every \(m,n\gt N_0\text{,}\) we have \(|a_m-a_n| \lt \epsilon/2\text{.}\) My the convergence of the subsequence, there exists a natural number \(N_1 \in \NN\) such that for every \(n_k\gt N_1\text{,}\) we have \(|a_{n_k}-L|\lt \epsilon/2\text{.}\) Now by the triangle inequality, if \(N=\max\{N_0,N_1\}\text{,}\) then for every \(n \gt N\text{,}\) we have \(|a_n-L| \lt \epsilon\text{.}\) This proves our claim.
Now assume that \((a_n)\) converges to a limit \(L\text{.}\) Let's prove that it is Cauchy. Let \(\epsilon \gt 0\text{.}\) By convergence, we may find a natural number \(N\) such that for every \(n\gt N\text{,}\) we have \(|a_n-L|\lt \epsilon/2\text{.}\) Now by the triangle inequality, if \(m,n \gt N\text{,}\) then \(|a_m-a_n|\lt \epsilon\text{.}\)