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Cauchy Sequence — Definition, Formula & Examples

A Cauchy sequence is a sequence whose terms become arbitrarily close to one another as you go far enough along. Instead of requiring a known limit, the definition only refers to the distances between terms themselves.

A sequence (an)(a_n) in a metric space (X,d)(X, d) is called a Cauchy sequence if for every ε>0\varepsilon > 0 there exists a positive integer NN such that for all m,n≥Nm, n \geq N, d(am,an)<εd(a_m, a_n) < \varepsilon. In R\mathbb{R} with the standard metric, this becomes ∣am−an∣<ε|a_m - a_n| < \varepsilon.

Key Formula

∀ ε>0,  ∃ N∈N   such that   m,n≥N  ⟹  ∣am−an∣<ε\forall\, \varepsilon > 0,\; \exists\, N \in \mathbb{N} \;\text{ such that }\; m, n \geq N \implies |a_m - a_n| < \varepsilon
Where:
  • ana_n = The $n$-th term of the sequence
  • ε\varepsilon = Any positive real number (the tolerance)
  • NN = An index beyond which all pairs of terms are within $\varepsilon$

How It Works

To show a sequence is Cauchy, you pick an arbitrary ε>0\varepsilon > 0 and then find an index NN beyond which every pair of terms is within ε\varepsilon of each other. You never need to identify the limit explicitly — the criterion is purely internal to the sequence. In R\mathbb{R}, every Cauchy sequence converges and every convergent sequence is Cauchy; this equivalence is precisely what it means for R\mathbb{R} to be complete. In incomplete spaces like Q\mathbb{Q}, a Cauchy sequence can fail to converge within the space.

Worked Example

Problem: Show that the sequence an=1na_n = \frac{1}{n} is a Cauchy sequence in R\mathbb{R}.
Step 1: Let ε>0\varepsilon > 0 be given. Estimate ∣am−an∣|a_m - a_n| for m,n≥Nm, n \geq N.
∣1m−1n∣=∣n−m∣mn\left|\frac{1}{m} - \frac{1}{n}\right| = \frac{|n - m|}{mn}
Step 2: Since m,n≥Nm, n \geq N, each reciprocal is at most 1N\frac{1}{N}. Apply the triangle inequality.
∣1m−1n∣≤1m+1n≤1N+1N=2N\left|\frac{1}{m} - \frac{1}{n}\right| \leq \frac{1}{m} + \frac{1}{n} \leq \frac{1}{N} + \frac{1}{N} = \frac{2}{N}
Step 3: Choose NN large enough so that 2N<ε\frac{2}{N} < \varepsilon, i.e., N>2εN > \frac{2}{\varepsilon}. Then for all m,n≥Nm, n \geq N, the Cauchy condition holds.
N>2ε  ⟹  ∣am−an∣<εN > \frac{2}{\varepsilon} \implies |a_m - a_n| < \varepsilon
Answer: For any ε>0\varepsilon > 0, choosing N>2εN > \frac{2}{\varepsilon} ensures ∣am−an∣<ε|a_m - a_n| < \varepsilon for all m,n≥Nm, n \geq N, so (1/n)(1/n) is Cauchy.

Why It Matters

Cauchy sequences are central to real analysis because they let you discuss convergence without knowing the limit in advance. The completeness of R\mathbb{R} — the fact that every Cauchy sequence of real numbers converges — underpins the rigorous construction of the reals from Q\mathbb{Q} and is used throughout measure theory, functional analysis, and differential equations.

Common Mistakes

Mistake: Assuming every Cauchy sequence converges regardless of the space it lives in.
Correction: Cauchy sequences converge only in complete spaces. For example, an=(1+1/n)na_n = (1 + 1/n)^n interpreted in certain incomplete subsets of R\mathbb{R} may not have a limit within that subset. In Q\mathbb{Q}, a sequence approaching 2\sqrt{2} is Cauchy but does not converge in Q\mathbb{Q}.