Show that a sequence converges if and only if it is a Cauchy sequence.

Steps to Solve

1. Understanding Convergence and Cauchy Sequences

A sequence ${a_n}$ converges to a limit $L$ if for every $\epsilon > 0$, there exists a natural number $N$ such that for all $n \geq N$, $|a_n - L| < \epsilon$.

A sequence ${a_n}$ is a Cauchy sequence if for every $\epsilon > 0$, there exists a natural number $N$ such that for all $m, n \geq N$, $|a_m - a_n| < \epsilon$.

2. Proving that Convergence Implies Cauchy

Assume ${a_n}$ converges to $L$.

Then for any $\epsilon > 0$, there is an $N_1$ such that $|a_n - L| < \frac{\epsilon}{2}$ for all $n \geq N_1$.

For any $m, n \geq N_1$, we have:

$$ |a_m - a_n| \leq |a_m - L| + |L - a_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$

Thus, ${a_n}$ is Cauchy.

3. Proving that Cauchy Implications Convergence

Now assume ${a_n}$ is a Cauchy sequence.

For every $\epsilon > 0$, there exists an $N_2$ such that $|a_m - a_n| < \epsilon$ for all $m, n \geq N_2$.

Since Cauchy sequences are bounded, we can find a limit $L$ such that the sequence converges to this limit.

4. Converging to a Limit

For all $n \geq N_2$, we can extract a convergent subsequence that will converge to some limit $L$. Since the whole sequence is Cauchy, it converges to $L$.

Thus, ${a_n}$ converges.