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

Steps to Solve

1. Definitions of Convergence and Cauchy sequences

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

2. Proving Convergence Implies Cauchy

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

   Given $\epsilon > 0$, there exists $N$ such that for all $n \geq N$, $|a_n - L| < \frac{\epsilon}{2}$.

   For $m, n \geq N$, we can use the triangle inequality:

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

   Thus, the sequence is Cauchy.

3. Proving Cauchy Implies Convergence

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

   Since $\mathbb{R}$ is complete, every Cauchy sequence converges to some limit $L$ in $\mathbb{R}$.

   For any $\epsilon > 0$, there exists $M$ such that for all $m, n \geq M$, we have $|a_m - a_n| < \epsilon$.

   Choose $k \geq M$, and we note that ${a_n}$ is bounded, and thus it converges to some limit $L$.

4. Conclusion

   We have shown both implications:

   • If a sequence converges, then it is a Cauchy sequence.
   • If a sequence is Cauchy, then it converges.

   Therefore, a sequence converges if and only if it is a Cauchy sequence.