Cauchy Sequences
Terms that crowd together — and why that means convergence in the reals.
The idea
The Cauchy criterion tests a sequence for convergence without reference to its limit. Verifying the definition of $a_{n} \to L$ requires knowing $L$ in advance; Cauchy's condition never mentions $L$.
Definition (Cauchy sequence).
A sequence $(a_{n})$ is Cauchy if $\forall \varepsilon > 0 \ \exists N \ \forall m, n \ge N : \ |a_{m} - a_{n}| < \varepsilon.$
Set this beside the definition of convergence: the quantifiers $\forall \varepsilon \ \exists N$ stand unchanged, but the single index $n$ has become a pair $m, n$, and the final clause compares $a_{m}$ with $a_{n}$ where convergence compared $a_{n}$ with $L$. The fixed target is gone; the terms are required to crowd around one another instead, so everything the statement mentions is a term of the sequence itself. Past $N$, all the terms fit inside a single band of width $\varepsilon$, anchored to the terms themselves rather than to any limit.
A convergent sequence is Cauchy: past some point every term lies within $\tfrac{\varepsilon}{2}$ of $L$, so any two such terms lie within $\varepsilon$ of each other.
The converse is a fact about the space the sequence lives in, not about the sequence. A space in which every Cauchy sequence converges to a point of that space is complete. The real line is complete; the rationals are not, since rational terms can crowd around an irrational point, leaving the sequence Cauchy in $\mathbb{Q}$ with no limit in $\mathbb{Q}$.
Theorem (Cauchy criterion).
A sequence of real numbers converges if and only if it is Cauchy.
In $\mathbb{R}$, therefore, Cauchy and convergent describe the same sequences, and we can certify convergence without ever producing the limit.
Ways to work on it
- Walkthrough. The Cauchy condition, a non-example, and finding an N.
- Practice. Decide whether a sequence is Cauchy.
- Hardest. A Cauchy sequence in the rationals with an irrational limit.
Not sure where to start? Take the ten-question placement test.