Cauchy Criterion (2.6.4)] A sequence converges if and only if it is a Cauchy sequence.


For the absolute value function definition and other properties see here.
For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template and an example, see this.

Discussion:

We already did the direction every convergent sequence is a Cauchy sequence in here. For the second part where we want to prove that a Cauchy sequence converges, we want to find \(N\) such that

$$ \begin{align*} |a_n - l| \leq \epsilon \quad \text{whenever $n > N$}. \end{align*} $$

But we don’t know what \(l\) is. How can we come up with such a candidate?? the trick is to use other theorems to get us there. In particular this proof uses the fact that every Cauchy sequence is bounded and then uses the Bolzano-Weierstrass theorem to conclude that there must exist a subsequence that is convergent from the original sequence. This limit is a good candidate because we also proved earlier that subsequences converge to the same limit as the original sequence in here)

So now we can go back and figure out what \(N\) would work here. We have \(|a_n - l|\). We can do the same exact trick of adding and subtracting the same term and then using the triangle inequality to come up with good candidate.

Proof:

The direction, a convergent sequence is a Cauchy sequence is proved here. For the other direction, we want to prove that if a sequence is Cauchy then it is convergent. Let \((a_n)\) be a Cauchy sequence. This means that \((a_n)\) is bounded by the proof we did earlier. And since \((a_n)\) is bounded, then by the Bolzano-Weierstrass theorem, we know that we must have a subsequence that is convergent. Let this subsequence be \((a_{n_k})\) and let \((a_{n_k}) \rightarrow l\). We will use \(l\) as the convergence candidate for \((a_n)\).

Since \((a_n)\) is Cauchy, then there exists \(N \in \mathbf{N}\) such that

$$ \begin{align*} |a_n - a_m| \leq \frac{\epsilon}{2} \quad \text{whenever $n, m > N$}. \end{align*} $$

We also know that \((a_{n_k}) \rightarrow l\). Let \(a_{n_K}\) be a term from the subsequence \((a_{n_k})\) such that \(n_K > N\). Therefore, the following holds

$$ \begin{align*} |a_{n_K} - l| \leq \frac{\epsilon}{2} \quad \text{whenever $n_K > N$}. \end{align*} $$

To prove that \((a_n) \rightarrow l\), we’ll prove that \(|a_n - l| \leq \epsilon\) whenever \(n > N\). To see that, we’ll add and subtract \(a_{n_K}\) and then use the triangular inequality to bound the term,

$$ \begin{align*} |a_n - l| &= |a_n + a_{n_K} - a_{n_K} - l| \\ &\leq |a_n - a_{n_K}| + |a_{n_K} - l| \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} $$

Therefore, \((a_n)\) converges to \(l\) as we wanted to show. \(\blacksquare\)

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