Cauchy Criterion for Convergence

The Cauchy Criterion provides a necessary and sufficient condition for convergence without reference to the limit. A sequence converges if and only if it is Cauchy: its terms eventually become arbitrarily close to each other. This criterion is fundamental in analysis and characterizes completeness of $\mathbb{R}$ .

Statement

Theorem2.1Cauchy Criterion

A sequence $(a_n)$ in $\mathbb{R}$ converges if and only if it is Cauchy. That is, $a_n \to L$ for some $L \in \mathbb{R}$ if and only if for every $\epsilon > 0$ , there exists $N \in \mathbb{N}$ such that

$|a_m - a_n| < \epsilon \quad \text{for all } m, n \geq N.$

Proof

Proof(⇒) Convergent implies Cauchy

Suppose $a_n \to L$ . Given $\epsilon > 0$ , choose $N$ such that $|a_n - L| < \epsilon/2$ for all $n \geq N$ . Then for $m, n \geq N$ ,

$|a_m - a_n| = |(a_m - L) + (L - a_n)| \leq |a_m - L| + |L - a_n| < \epsilon/2 + \epsilon/2 = \epsilon.$

Thus $(a_n)$ is Cauchy.

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Proof(⇐) Cauchy implies convergent

Suppose $(a_n)$ is Cauchy. We show it converges.

Step 1: $(a_n)$ is bounded. Choose $N$ such that $|a_m - a_n| < 1$ for all $m, n \geq N$ . Then $|a_n - a_N| < 1$ for $n \geq N$ , so $|a_n| < |a_N| + 1$ for $n \geq N$ . Let $M = \max(|a_1|, \ldots, |a_{N-1}|, |a_N| + 1)$ . Then $|a_n| \leq M$ for all $n$ .

Step 2: $(a_n)$ has a convergent subsequence. By the Bolzano-Weierstrass theorem, the bounded sequence $(a_n)$ has a convergent subsequence $a_{n_k} \to L$ .

Step 3: The full sequence $a_n \to L$ . Given $\epsilon > 0$ , choose $N_1$ such that $|a_m - a_n| < \epsilon/2$ for all $m, n \geq N_1$ (Cauchy condition). Choose $K$ such that $n_K \geq N_1$ and $|a_{n_K} - L| < \epsilon/2$ (subsequence converges to $L$ ). Then for $n \geq N_1$ ,

$|a_n - L| = |(a_n - a_{n_K}) + (a_{n_K} - L)| \leq |a_n - a_{n_K}| + |a_{n_K} - L| < \epsilon/2 + \epsilon/2 = \epsilon.$

Thus $a_n \to L$ .

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RemarkCompleteness

The direction (⇐) crucially uses the completeness of $\mathbb{R}$ via the Bolzano-Weierstrass theorem. Over $\mathbb{Q}$ , Cauchy sequences need not converge (e.g., decimal approximations to $\sqrt{2}$ ).

Applications

ExampleCauchy criterion for series

A series $\sum a_n$ converges if and only if the partial sums $s_n = \sum_{k=1}^n a_k$ form a Cauchy sequence. Equivalently: for every $\epsilon > 0$ , there exists $N$ such that

$\left|\sum_{k=m}^n a_k\right| < \epsilon \quad \text{for all } n \geq m \geq N.$

This is the Cauchy criterion for series and is used to prove convergence tests (comparison, ratio, root tests).

ExampleConvergence of e^x series

The series $e^x = \sum_{n=0}^\infty x^n/n!$ converges for all $x \in \mathbb{R}$ . To see this, for $m < n$ ,

$\left|\sum_{k=m}^n \frac{x^k}{k!}\right| \leq \sum_{k=m}^n \frac{|x|^k}{k!} \to 0 \quad \text{as } m \to \infty$

(tail of a convergent series). By the Cauchy criterion, the partial sums converge.

ExampleFailure in Q

Let $a_n$ be the $n$ -th decimal approximation to $\sqrt{2}$ : $a_1 = 1$ , $a_2 = 1.4$ , $a_3 = 1.41$ , etc. Then $(a_n)$ is Cauchy as a sequence in $\mathbb{Q}$ , but does not converge in $\mathbb{Q}$ (since $\sqrt{2} \notin \mathbb{Q}$ ). This shows $\mathbb{Q}$ is not complete.

Summary

The Cauchy criterion characterizes convergence intrinsically:

A sequence converges iff it is Cauchy.
Equivalently: $\mathbb{R}$ is complete (all Cauchy sequences converge).
Applications include convergence tests for series and uniform convergence.

See Bolzano-Weierstrass for the key lemma used in the proof, and Metric Spaces for generalizations.