Theorem
A sequence converges if and only if it is a Cauchy sequence.

Strategy

We already did the direction every convergent sequence is a Cauchy sequence in here. For the second part where we want to prove that if we have a Cauchy sequence, then it convergences to some limit. I.e. 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 the limit?? the trick is to use the fact that every Cauchy sequence is bounded. One we show it’s bounded, then we can use Bolzano-Weierstrass to conclude that there must exist a subsequence that is convergent from the original sequence. At the same time, we know that subsequences converge to the same limit as the original sequence. Hence, we are done.

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. 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 \(\{a_{n_k}\} \rightarrow \ell\). We will show that \(\{a_n\}\) converges to \(\ell\).

Now, since \(\{a_n\}\) is Cauchy, then there exists \(N_0 \in \mathbb{N}\) such that whenever \(n,m \geq N_0\),

$$ \begin{align*} |a_n - a_m| \leq \frac{\epsilon}{2} \end{align*} $$

Since \(a_{n_k}\to \ell\), there exists \(K_1\in\mathbb{N}\) such that if \(k\ge K_1\), then

$$ \begin{align*} |a_{n_k} - \ell| < \frac{\epsilon}{2} \end{align*} $$

Since \(n_k\to\infty\), there exists \(K_2\in\mathbb{N}\) such that if \(k\ge K_2\), then \(n_k\ge N_0\). Choose \(K \ge \max\{K_1,K_2\}\) and set \(N=N_0\). Then for any \(n\ge N\) we have \(n\ge N_0\) and \(n_K\ge N_0\), so

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

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

References