Every Convergent Sequence is Cauchy

Real Analysis · Axiom Academy

EXAMPLE Every Convergent Sequence is Cauchy Proving that if then (a_n) is Cauchy, via the triangle inequality. Prove that every convergent sequence is a Cauchy sequence. That is, show that if , then for every there exists N so that whenever . Recall. (a_n) converges to L if for all there is an N with for all . (a_n) is Cauchy if for all there is an N with for all . Once the index passes N , every term lands inside the shaded band of half-width ε/2 around L. Two such terms can be no more than ε apart — that gap is exactly the Cauchy condition. Each green term is within ε/2 of the red limit L, so the worst-case separation between them is ε/2 + ε/2 = ε. The triangle inequality makes that bound rigorous: it lets us measure the gap between the two terms by routing through L. Nicely done — you proved that convergence forces the Cauchy condition. The whole proof rides on one tactic: ask convergence for half the room you ultimately need. The ε/2 trick: bounding each of two pieces by makes their sum land below — the standard move whenever a quantity splits into two errors. Triangle inequality routes through L : trades the unknown gap between terms for two known distances to the limit. Result: for any , taking N from convergence at tolerance gives for all , so (a_n) is Cauchy. One-way in general: every convergent sequence is Cauchy, but the converse needs completeness — in Cauchy sequences do converge, yet in they need not.

This is the written version of the interactive lesson above. See the full Real Analysis course.