Cauchy Criterion

Theorem

A sequence converges iff it is a Cauchy Sequence.

Proof

( $\Rightarrow$ ): See Cauchy Sequences Converge.

( $\Leftarrow$ ): Start with the Cauchy sequence $(x_{n})$ . Using Cauchy Sequences Are Bounded then $(x_{n})$ is bounded. We can use the Balzano-Weierstrass Theorem to produce a convergent subsequence $(x_{n_{k}})$ . Set:

x = lim_{n \to \infty} x_{n_{k}}

The idea is to show that $x_{n} \to x$ . We'll use the triangle inequality for this argument. We know the terms in the subsequence are getting close to the limit $x$ , and the assumption that $(x_{n})$ is Cauchy implies the terms in the "tail" of the sequence are close to each other. Thus, we want to make each of these distances less than half of the prescribed $ϵ$ .

Let $ϵ > 0$ . Because ( $x_{n}$ ) is Cauchy then $\exists N$ where:

| x_{n} - x_{m} | < \frac{ϵ}{2}

whenever $m, n \geq N$ . Now we also know $(x_{n_{k}}) \to x$ so choose a term in this subsequence, denoted $x_{n_{K}}$ with $n_{K} \geq N$ and:

| x_{n_{K}} - x | < \frac{ϵ}{2}

To see that $N$ has the desired property for the original sequence $(x_{n})$ observe that any $n > N$ has:

\begin{aligned} | x_{n} - x | & = | x_{n} - x_{n_{K}} + x_{n_{K}} - x | \\ \leq | x_{n} - x_{n_{K}} | + | x_{n_{K}} - x | \\ < \frac{ϵ}{2} + \frac{ϵ}{2} \\ = ϵ \end{aligned}

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