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Lesson 7 of 20

Cauchy sequences and completeness

Control pairs of tail terms and locate exactly where completeness is used.

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For a_n=1/n, late terms are close to one another even before naming their limit. If m,n≥N, then |1/m−1/n|≤1/N. A Cauchy sequence makes this pairwise tail distance arbitrarily small. The definition is useful when a limit is difficult to identify: it asks about known terms rather than an unknown destination. It still requires control over every pair of sufficiently late indices.

Predict

Predict whether checking neighboring terms is enough. For harmonic partial sums H_n=1+1/2+⋯+1/n, the difference H_(n+1)−H_n is small. But H_(2n)−H_n is the sum of n terms, each at least 1/(2n), so it is at least 1/2. Nearby changes shrink while long-tail changes remain substantial. This distinction is why a Cauchy condition quantifies over independent m and n.

Experiment

Compare the quantifiers explicitly.

PropertyRequired comparison
Successive differences vanishFor all late n, compare a_(n+1) with a_n
CauchyFor all late m,n, compare a_m with a_n
Convergent to LFor all late n, compare a_n with fixed L

For a chosen tolerance 1/4, use the harmonic block estimate to defeat any proposed Cauchy cutoff. Pick n beyond that cutoff and m=2n. Both indices are late, yet their terms remain too far apart.

Understand

Every convergent sequence is Cauchy. Given ε>0, use convergence to put both a_m and a_n within ε/2 of L beyond one cutoff. Their mutual distance is less than ε by the triangle inequality.

Conversely, begin with a real Cauchy sequence. Tolerance 1 bounds its tail relative to a fixed term a_N; the finite prefix is bounded separately. Bolzano–Weierstrass therefore gives a subsequence converging to some real L. For an arbitrary ε, choose a Cauchy cutoff with pairwise distances below ε/2. Choose one subsequence term beyond that cutoff and within ε/2 of L. Every term in the Cauchy tail is within ε/2 of this selected term, hence within ε of L. This proves convergence of the whole sequence, not just the extraction.

Apply

The reverse implication uses completeness through Bolzano–Weierstrass. It is false if the ambient space is Q. Rational approximations to √2 can be Cauchy because their pairwise distances shrink, yet their only possible real limit is not rational. The distances alone do not force the missing point into the smaller space.

Repair “Cauchy means the terms eventually become equal.” Equality is much stronger. The distinct values 1/n form a Cauchy sequence. The definition concerns all positive tolerances, with a potentially different cutoff for each; it never requires one finite stage after which every distance is exactly zero.

Continue

Optional depth connects this theorem to construction of the real numbers from rational Cauchy sequences. That construction reverses the logical starting point: instead of assuming least upper bounds, it builds a complete ordered field and then proves the corresponding order property. One must not silently mix the two foundations in a circular proof.

Save a proof of each implication and highlight where the real-number hypothesis enters. Then write a non-Cauchy witness as a fixed tolerance plus a rule producing two late indices. This tail-control language will transfer directly to infinite series through their sequences of partial sums.

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Practice and reasoning

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For a_n=1/n, calculate |a_4−a_8|.

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A rational Cauchy sequence necessarily:

Your response

Prove a convergent sequence is Cauchy.

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Prove a Cauchy sequence is bounded.

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Prove a real Cauchy sequence converges using Bolzano–Weierstrass.

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Disprove: |a_(n+1)−a_n|→0 implies the sequence is Cauchy.

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Reflect and continue

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