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

Steps to Solve

1. Define Convergence and Cauchy Sequence

A sequence ${x_n}$ converges to a limit $L$ if for every $\epsilon > 0$, there exists a positive integer $N$ such that for all $n \geq N$, the terms of the sequence satisfy: $$ |x_n - L| < \epsilon $$

A sequence is called a Cauchy sequence if for every $\epsilon > 0$, there exists a positive integer $M$ such that for all $m, n \geq M$, the terms satisfy: $$ |x_n - x_m| < \epsilon $$

2. Show Convergence Implies Cauchy Sequence

Assume ${x_n}$ converges to $L$. Given any $\epsilon > 0$, by the definition of convergence, choose $N$ such that for all $n \geq N$, we have: $$ |x_n - L| < \frac{\epsilon}{2} $$

Now for all $m, n \geq N$: [ |x_n - x_m| \leq |x_n - L| + |L - x_m| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon ] Thus, ${x_n}$ is a Cauchy sequence.

3. Show Cauchy Sequence Implies Convergence

Now assume ${x_n}$ is a Cauchy sequence. We must show that it converges. Since the real numbers are complete, every Cauchy sequence converges to a limit $L$.

To show this explicitly, note that as $n$ increases, the distances between the terms $x_n$ become arbitrarily small. Therefore, we can find a limit point $L$ such that for any $\epsilon > 0$, there exists $M$ such that for all $m, n \geq M$: $$ |x_n - x_m| < \epsilon $$

Taking one of the terms, say $x_M$, we can show that $x_n$ converges to this limit $L$.

4. Conclusion of the Proof

After establishing both directions:

• If ${x_n}$ converges to $L$, then it is a Cauchy sequence.
• If ${x_n}$ is a Cauchy sequence, then it converges to some limit $L$.

This proves that a sequence converges if and only if it is a Cauchy sequence.