Convergence names a limit. There is a weaker demand that names nothing: that the terms eventually get close to each other. A sequence whose terms huddle together like this is called Cauchy, and it is what you can check when the limit is exactly the thing you are trying to find.

A search party closing in reports that its members are all within fifty metres of each other. That is progress, and it is checkable on the radio — without anyone knowing yet whether there is anything at the middle to find.

The sequence $1/n$ inside the space $(0,1]$, where the point $0$ is missing. The terms crowd together — beyond the tenth they are all within $0.1$ of each other — and there is nothing in the space for them to crowd around. Being Cauchy is about the terms; converging is about the space as well.

A sequence in a metric space is Cauchy when for every tolerance there is an index beyond which any two terms are within that tolerance of each other. The limit does not appear: the condition compares terms with terms.

• Every convergent sequence is Cauchy, which the theorem below proves. The converse is the interesting direction, and it depends on the space.
• In $(0,1]$ the sequence $1/n$ is Cauchy and converges to nothing: its only candidate limit, $0$, is not a point of the space. In $\mathbb{Q}$ the sequence $1,1.4,$ is Cauchy and has no rational limit.

Given a tolerance, apply the definition of convergence to half of it. Beyond the index it provides, every term is within half the tolerance of the limit. Two such terms are then within a tolerance of each other, by going from one to the other through the limit. Halving first is what makes the triangle inequality come out right, and it is the standard move.

Take any tolerance and halve it; half of a positive number is again a tolerance the definition of convergence must meet.

Convergence gives an index beyond which every term is that close to the limit.

Take any two indices beyond it; the claim is about this pair.

Travel from one term to the other through the limit and use the two bounds.

The two halves make one tolerance, which is what being Cauchy asks for.

The Cauchy condition compares two late terms with each other. No limit appears in it, which is why it can be checked without knowing one.

For $x_n=1+\tfrac12+\cdots+\tfrac1n$, consecutive terms are as close as you like: .

• A sequence is Cauchy when its terms eventually stay within any tolerance of each other.
• Convergent sequences are Cauchy; the converse depends on the space.
• Consecutive gaps tending to $0$ is not enough — remember $1+\tfrac12+\cdots+\tfrac1n$.

Great work! You now know how to:

• state the Cauchy condition and tell it from convergence;
• prove that a convergent sequence is Cauchy;
• give a Cauchy sequence that converges to nothing;
• reject the argument that consecutive terms closing up is enough.