Sequence limits and epsilon–N proofs

Try

The sequence a_n=3+2/n has values 5,4,11/3 and so on, suggesting a limit of 3. A plot or table can show a long finite prefix, but convergence is a claim about every sufficiently late index. For a desired error ε, solve 2/n<ε. Any integer N>2/ε works for all n≥N. The algebra produces the whole-tail guarantee that a picture alone cannot provide.

Predict

Predict which choice comes first: the tolerance ε or the cutoff N. The definition lets an adversary specify any ε>0, after which the proof chooses N. That single N must control every later n. Choosing a different cutoff after seeing each individual n would not establish convergence. The candidate limit is also fixed before the tolerance challenge; it cannot drift with the requested accuracy.

Experiment

Make the dependence visible.

ε	One valid N	All n≥N satisfy
1	3	|a_n−3|<1
1/10	21	|a_n−3|<1/10
1/100	201	|a_n−3|<1/100

These choices are convenient, not uniquely required. A larger cutoff remains valid. Explain why the first several terms may violate a tolerance without affecting convergence. Then change the numerator from 2 to 7 and update the rule rather than manually listing hundreds of terms.

Understand

Formally, a_n→L means: for every ε>0 there exists N∈N such that every n≥N satisfies |a_n−L|<ε. The quantifiers express uniform control over a tail of the sequence.

A limit is unique. Suppose L and M are distinct limits, and put d=|L−M|>0. Use tolerance d/3 for each convergence claim and take the larger cutoff. Beyond it, the triangle inequality gives d≤|L−a_n|+|a_n−M|<2d/3, a contradiction.

Every convergent sequence is bounded. Tolerance 1 bounds the tail by |L|+1. The finitely many earlier terms have a maximum absolute value; the larger of that maximum and the tail bound controls all terms. A proof that ignores the prefix proves eventual boundedness but has not yet justified the full statement.

Apply

To negate convergence to a proposed L, find ε₀>0 such that for every N some n≥N has |a_n−L|≥ε₀. Notice that the violating n may depend on N, while ε₀ stays fixed. The sequence (−1)^n has even terms 1 and odd terms −1 arbitrarily far out. No single L can be within 1/2 of both values, so it does not converge.

Repair “the successive differences get small, so the sequence converges.” Harmonic partial sums have differences 1/(n+1) tending to zero but keep growing. Small neighboring changes do not control cumulative change across a long tail. A later Cauchy criterion will compare all pairs of late terms.

Continue

Optional depth compares explicit convergence rates. The estimate 2/n gives a direct tolerance-to-index recipe, while an abstract existence proof may establish convergence without a practical numerical cutoff. Both are mathematically valid, but they support different computational claims.

Write one complete proof beginning “Let ε>0” and ending with an inequality valid for every n≥N. Separately write its negation, preserving the quantifier order. The next lesson uses this language to prove sum, product and quotient laws instead of treating them as rules justified only by successful numerical examples.