Convergent sequences and the ε–N game

What does it mean for a sequence of numbers to converge to a limit L?

Someone names a band of half-height ε around L. You answer with an index N such that every term from the N-th on lies inside the band. The sequence converges to L when you can answer every ε > 0, however small.

A row of twelve poles, each topped with a gold ball, in front of two horizontal ropes. Balls 1 and 3 are above the upper rope and balls 2 and 4 below the lower rope. Balls 5 to 12 are all between the ropes, alternately a little above and a little below the middle. — Pole \(n\) stands for the term \(a_n\): its ball sits at height \(a_n\). The two ropes mark the band from \(L - \varepsilon\) to \(L + \varepsilon\). From pole 5 on, every ball in the picture is strictly inside the band, so \(N = 5\) works for the 12 terms shown. The definition asks the same of every later term, and for every \(\varepsilon \gt 0\).

Before you start · 3 quick checks on absolute value, strict inequalities and the order of “for every” and “there is”, about 3 min

Definitions

A sequence is a list of real numbers \(a_1, a_2, a_3, \ldots\), one term \(a_n\) for each whole number \(n \ge 1\). Formally it is a function from the whole numbers \(1, 2, 3, \ldots\) to the real numbers. On this page “whole number” means one of \(1, 2, 3, \ldots\).

Limit of a sequence. The sequence \((a_n)\) converges to the number \(L\), written \(a_n \to L\) or \(\lim_{n \to \infty} a_n = L\), when for every \(\varepsilon \gt 0\) there is a whole number \(N\) such that

\[ |a_n - L| \lt \varepsilon \quad \text{for every } n \ge N. \]

\(|a_n - L| \lt \varepsilon\) means \(L - \varepsilon \lt a_n \lt L + \varepsilon\): the term lies in the band of half-height \(\varepsilon\) around \(L\). A term with \(|a_n - L| = \varepsilon\) sits on the edge and is not in the band. A sequence that converges to some number is convergent; one that converges to no number is divergent.

The Archimedean property. For every real number \(x\) there is a whole number \(N\) with \(N \gt x\): the whole numbers have no upper bound.

Fine print: what N may depend on, and what does not matter

\(N\) is chosen after \(\varepsilon\) and may depend on it. It need not be the smallest index that works: if \(N\) works, so does every larger whole number.

Only the tail matters: changing finitely many terms changes neither whether the sequence converges nor its limit. Some sequences start at \(n = 0\) or at another index; the definition is the same.

The lesson on why the real numbers can't be listed uses lists \(a_1, a_2, a_3, \ldots\) of this kind and shows that no sequence contains every real number.

1 See

Defend the limit: answer ε with N

In each goal the page names a sequence, a claimed limit \(L\) and a band half-height \(\varepsilon\). You type a whole number \(N\) and press Defend. The page then checks the terms \(a_N, a_{N+1}, \ldots\) and shows the first one outside the band, if there is one. If you think no \(N\) can work, press No N works.

Your N, a whole number from 1 to 80

How the page checks every term from N on

2 Try

Switch sides: you name ε

Two questions first. The challenge below opens when you have answered both.

The challenge opens when you have answered the two questions above.

3 Prove

A first ε–N proof

Theorem

Let \(a_n = \dfrac{n^2 + n}{2n^2 + 1}\) for whole numbers \(n \ge 1\). Then \(a_n \to \tfrac12\): for every \(\varepsilon \gt 0\) there is a whole number \(N\) such that \(|a_n - \tfrac12| \lt \varepsilon\) for every \(n \ge N\).

Facts the proof uses.

the Archimedean property from the setup;
for positive numbers, a smaller numerator or a larger denominator gives a smaller fraction: if \(0 \lt p \le p'\) and \(q \ge q' \gt 0\), then \(\tfrac{p}{q} \le \tfrac{p'}{q'}\), with strict inequality when \(p \lt p'\) or \(q \gt q'\).

Build the proof one line at a time; each line has one correct choice. Each option is a statement about the \(\varepsilon\), \(N\) and \(n\) of the proof. A wrong option is false in every case it speaks of, or claims a rule that fails.

What follows

A sequence has at most one limit

Suppose \(a_n \to L\) and \(a_n \to M\) with \(L \ne M\). Take \(\varepsilon = \tfrac{|L - M|}{2}\), which is positive. There is \(N_1\) with \(|a_n - L| \lt \varepsilon\) for every \(n \ge N_1\), and \(N_2\) with \(|a_n - M| \lt \varepsilon\) for every \(n \ge N_2\). For \(n = \max(N_1, N_2)\) both hold, and the triangle inequality gives

\[ \begin{aligned} |L - M| &\le |L - a_n| + |a_n - M| \\ &\lt \varepsilon + \varepsilon = |L - M|, \end{aligned} \]

which is impossible. So \(L = M\).

The squeeze theorem

Suppose \(b_n \le a_n \le c_n\) for every \(n \ge N_0\), and \(b_n \to L\) and \(c_n \to L\). Then \(a_n \to L\).

Proof. Let \(\varepsilon \gt 0\). There is \(N_1\) with \(L - \varepsilon \lt b_n\) for every \(n \ge N_1\), and \(N_2\) with \(c_n \lt L + \varepsilon\) for every \(n \ge N_2\). For every \(n \ge N = \max(N_0, N_1, N_2)\), \(L - \varepsilon \lt b_n \le a_n \le c_n \lt L + \varepsilon\), so \(|a_n - L| \lt \varepsilon\).

For example, \(-\tfrac{1}{n} \le \tfrac{\sin n}{n} \le \tfrac{1}{n}\) for every \(n \ge 1\) (with \(n\) in radians), and both bounds tend to 0, so \(\tfrac{\sin n}{n} \to 0\).

Where this leads

The ε–δ definition of a limit plays the same game for a function: a window \(0 \lt |x - a| \lt \delta\) takes the place of the tail \(n \ge N\). The two are linked: for \(f: D \to \mathbb{R}\) and a limit point \(a\) of \(D\), \(f(x) \to L\) as \(x \to a\) exactly when \(f(x_n) \to L\) for every sequence \((x_n)\) in \(D\) with \(x_n \ne a\) and \(x_n \to a\).

The lesson on where e comes from uses the monotone convergence theorem: an increasing sequence whose terms stay below a fixed number converges. Its proof needs the completeness of the real numbers (every nonempty set of real numbers with an upper bound has a least upper bound), which the ε–N definition alone does not give.

4 Apply

Problems

Each problem has hints and a full solution. Answers that ask for \(N\) are whole numbers and are graded exactly; type digits only.

Key point To find the smallest \(N\), write \(|a_n - L|\) as a formula in \(n\), solve the inequality \(|a_n - L| \lt \varepsilon\), and check that every later term also satisfies it (here because the distance decreases). To show that a sequence does not converge to \(L\), find one \(\varepsilon\) that defeats every \(N\).

Level Intro Prerequisites absolute value and inequalities Time ~35 min

Next: The ε–δ definition of a limit. Related: Where e comes from and Why the real numbers can't be listed.