Limits, proved

Calculus rests on the statement that a sequence converges to a limit, and so far that statement has been written down, read as a game and negated, but never proved for a single sequence.

The previous lesson supplied what was missing: the completeness axiom, that every nonempty set of reals bounded above has a least upper bound, and from it the Archimedean property, that for every real $x$ some natural number exceeds $x$ . This lesson puts the whole course to work on one definition. It proves that particular sequences converge and that others do not, proves general theorems about limits, and ends with a theorem that is true in $ℝ$ and false in $ℚ$ , which is exactly what completeness buys. Along the way it separates two activities that textbooks tend to blur: finding a proof and writing one.

The definition, read closely

A sequence $a_{1}, a_{2}, a_{3}, \dots$ of real numbers converges to $L$ when

\forall ε > 0 \exists N \in ℕ \forall n \geq N (| a_{n} - L | < ε)

with $ℕ = {1, 2, 3, \dots}$ as throughout the course. The lesson on quantifiers read it as a game. A sceptic names a tolerance $ε > 0$ , as small as they like. A prover answers with a position $N$ . The sceptic then picks any term from position $N$ on, and the prover wins if that term lies within $ε$ of $L$ . The sequence converges to $L$ when the prover has a winning answer to every tolerance, and we write $a_{n} \to L$ .

Because $\exists N$ comes after $\forall ε$ , the position may depend on the tolerance, and for any interesting sequence it must: a smaller $ε$ usually needs a later $N$ . Because the last quantifier is $\forall n \geq N$ , it is not enough for one term to land close; every term from $N$ on must, so the sequence stays close rather than visiting. The first $N - 1$ terms are ignored completely, which is why changing finitely many terms never changes a limit.

The quantifiers also dictate the shape of every convergence proof before any arithmetic is done. The universal $\forall ε > 0$ is met by "Let $ε > 0$ ", assuming nothing about it but positivity. The existential $\exists N$ is met by naming an $N$ , which may be built from $ε$ but from nothing chosen later. The inner universal is met by "Let $n \geq N$ ". What remains is to show $| a_{n} - L | < ε$ using only those facts. An $N$ that works may always be replaced by a larger one, so there is no need to find the smallest; the proof only has to find one.

The first limit: $\frac{1}{n} \to 0$

Theorem. The sequence $a_{n} = \frac{1}{n}$ converges to $0$ .

Proof. Let $ε > 0$ . By the Archimedean property there is a natural number $N$ with $N > \frac{1}{ε}$ . Let $n \geq N$ . Then

| a_{n} - 0 | = \frac{1}{n} \leq \frac{1}{N} < ε

where the first inequality holds because $n \geq N > 0$ , and the second follows from $N > \frac{1}{ε}$ on multiplying both sides by the positive number $\frac{ε}{N}$ . Since $ε$ was arbitrary, $\frac{1}{n} \to 0$ . $□$

For $ε = 0.01$ the proof asks for $N > 100$ , and $N = 101$ is the answer the lesson on quantifiers found by hand. The proof answers every tolerance at once, with a recipe rather than a table.

The one step that is not arithmetic is the existence of $N$ , and it is not a formality. It is the Archimedean property, which the previous lesson derived from completeness. There are ordered fields that obey every rule of school algebra and contain positive elements smaller than every $\frac{1}{n}$ ; in such a field the sceptic names one of those as $ε$ and wins. The most basic limit in analysis is already a fact about the reals, not about algebra.

How a proof is found, and how it is written

For most sequences the right $N$ is not obvious, and the finished proof hides how it was found, because the work happens in two passes. The first is scratch work, done on the side and never handed in: start from the goal $| a_{n} - L | < ε$ , simplify $| a_{n} - L |$ , and work backwards to a condition on $n$ that guarantees it. The second is the proof, which runs forwards: it starts from $ε$ , names the $N$ the scratch work found, and checks that it wins.

Example. Prove that $a_{n} = \frac{3 n + 1}{n + 2}$ converges to $3$ .

Scratch work. Simplify the distance to the limit:

| a_{n} - 3 | = | \frac{3 n + 1 - 3 (n + 2)}{n + 2} | = \frac{5}{n + 2}

The goal $\frac{5}{n + 2} < ε$ holds exactly when $n > \frac{5}{ε} - 2$ . That is the sharpest condition, but a cruder one is easier to write with: since $\frac{5}{n + 2} < \frac{5}{n}$ , it is enough that $\frac{5}{n} < ε$ , which holds once $n > \frac{5}{ε}$ . So take $N > \frac{5}{ε}$ .

Proof. Let $ε > 0$ . By the Archimedean property there is $N \in ℕ$ with $N > \frac{5}{ε}$ . Let $n \geq N$ . Then

| a_{n} - 3 | = \frac{5}{n + 2} < \frac{5}{n} \leq \frac{5}{N} < ε

so $| a_{n} - 3 | < ε$ for every $n \geq N$ . Since $ε$ was arbitrary, $a_{n} \to 3$ . $□$

At $ε = 0.1$ the proof asks for $N > 50$ , so $N = 51$ , while the sharp condition $n > 48$ shows the smallest working position is $49$ . Both win, and the proof is shorter for not caring which. What the proof never contains is the backwards reasoning: "suppose $| a_{n} - 3 | < ε$ , then..." would be assuming the conclusion, and a reader would rightly object.

Now you. Prove that $a_{n} = \frac{4 n - 1}{2 n + 3}$ converges to $2$ , showing the scratch work and the proof separately.

Answer

Scratch work. $| a_{n} - 2 | = | \frac{4 n - 1 - 2 (2 n + 3)}{2 n + 3} | = \frac{7}{2 n + 3}$ , which is less than $\frac{7}{2 n}$ . That is below $ε$ once $n > \frac{7}{2 ε}$ .

Proof. Let $ε > 0$ . By the Archimedean property there is $N \in ℕ$ with $N > \frac{7}{2 ε}$ . Let $n \geq N$ . Then $| a_{n} - 2 | = \frac{7}{2 n + 3} < \frac{7}{2 n} \leq \frac{7}{2 N} < ε$ . Since $ε$ was arbitrary, $a_{n} \to 2$ . $□$

Proving that a sequence diverges

A sequence diverges when it converges to no real number at all. The lesson on quantifiers negated the definition: $a_{n}$ fails to converge to $L$ when

\exists ε > 0 \forall N \in ℕ \exists n \geq N (| a_{n} - L | \geq ε)

The roles swap. Now the prover picks one tolerance, the sceptic picks any starting position, and the prover must find a term beyond it that misses. To prove divergence, this must be done for every candidate $L$ , so the proof starts "Let $L$ be any real number."

Example. Prove that $a_{n} = (- 1)^{n}$ diverges.

Let $L$ be any real number, and take $ε = 1$ . Let $N \in ℕ$ . The terms at positions $2 N$ and $2 N + 1$ are both beyond $N$ , and they are $1$ and $- 1$ . If both were within $1$ of $L$ , the triangle inequality would give

2 = | 1 - (- 1) | \leq | 1 - L | + | L - (- 1) | < 1 + 1 = 2

which is impossible. So one of the two positions gives an $n \geq N$ with $| a_{n} - L | \geq 1$ . Since $N$ was arbitrary, $a_{n}$ does not converge to $L$ , and since $L$ was arbitrary, it converges to nothing. $□$

The proof never says which of the two terms misses, only that one does, and an existential claim needs no more.

Now you. Prove that $a_{n} = n$ diverges.

Answer

Let $L$ be any real number, and take $ε = 1$ . Let $N \in ℕ$ . By the Archimedean property there is a natural number $m > L + 1$ ; let $n$ be the larger of $N$ and $m$ . Then $n \geq N$ and $n > L + 1$ , so $| a_{n} - L | \geq n - L > 1$ . Since $N$ and then $L$ were arbitrary, $a_{n}$ diverges. $□$

Uniqueness and the $\frac{ε}{2}$ argument

Writing $lim a_{n} = L$ presumes a sequence has only one limit, and that needs proof.

Theorem. If $a_{n} \to L$ and $a_{n} \to M$ , then $L = M$ .

Proof. Suppose not, so $L \neq M$ , and let $ε = \frac{| L - M |}{2}$ , which is positive. Since $a_{n} \to L$ there is $N_{1}$ with $| a_{n} - L | < ε$ for all $n \geq N_{1}$ , and since $a_{n} \to M$ there is $N_{2}$ with $| a_{n} - M | < ε$ for all $n \geq N_{2}$ . Let $n$ be the larger of $N_{1}$ and $N_{2}$ . By the triangle inequality,

| L - M | \leq | L - a_{n} | + | a_{n} - M | < ε + ε = | L - M |

a contradiction. So $L = M$ . $□$

The idea is to spend the tolerance in halves: two distances each below $\frac{ε}{2}$ sum to less than $ε$ , and taking the later of two positions makes both guarantees hold at once. The same budget proves that limits respect addition.

Example. Prove that if $a_{n} \to L$ and $b_{n} \to M$ , then $a_{n} + b_{n} \to L + M$ .

Let $ε > 0$ . Then $\frac{ε}{2} > 0$ , so, applying the definition to each sequence with tolerance $\frac{ε}{2}$ , there are $N_{1}$ and $N_{2}$ with $| a_{n} - L | < \frac{ε}{2}$ for $n \geq N_{1}$ and $| b_{n} - M | < \frac{ε}{2}$ for $n \geq N_{2}$ . Let $N$ be the larger of $N_{1}$ and $N_{2}$ , and let $n \geq N$ . Then

| (a_{n} + b_{n}) - (L + M) | \leq | a_{n} - L | + | b_{n} - M | < \frac{ε}{2} + \frac{ε}{2} = ε

Since $ε$ was arbitrary, $a_{n} + b_{n} \to L + M$ . $□$

Now you. Prove that if $a_{n} \to L$ , then $a_{n + 1} - a_{n} \to 0$ .

Answer

Let $ε > 0$ . There is $N$ with $| a_{n} - L | < \frac{ε}{2}$ for all $n \geq N$ . Let $n \geq N$ ; then $n + 1 \geq N$ too, so

| a_{n + 1} - a_{n} | \leq | a_{n + 1} - L | + | L - a_{n} | < \frac{ε}{2} + \frac{ε}{2} = ε

Since $ε$ was arbitrary, $a_{n + 1} - a_{n} \to 0$ . $□$ For $(- 1)^{n}$ the consecutive differences are always $\pm 2$ , which gives a second proof that it diverges.

Bounded, increasing, and what completeness buys

A sequence is bounded when some $B$ has $| a_{n} | \leq B$ for every $n$ . Every convergent sequence is. If $a_{n} \to L$ , the definition with $ε = 1$ gives $N$ with $| a_{n} - L | < 1$ for $n \geq N$ , and then $| a_{n} | \leq | a_{n} - L | + | L | < | L | + 1$ . Only the finitely many terms before position $N$ are left, so $B$ , the largest of $| a_{1} |, \dots, | a_{N - 1} |$ and $| L | + 1$ , bounds them all. The converse fails, as the bounded divergent $(- 1)^{n}$ shows. But one extra condition rescues it.

A sequence is increasing when $a_{n} \leq a_{n + 1}$ for every $n$ , and then, by induction, $a_{n} \leq a_{m}$ whenever $n \leq m$ .

Theorem (monotone convergence). A bounded increasing sequence of reals converges to the supremum of its terms.

Proof. The set $A = {a_{n} : n \in ℕ}$ is nonempty and bounded above, so by the completeness axiom it has a supremum $s$ . Let $ε > 0$ . Since $s - ε < s$ , it is not an upper bound of $A$ (the supremum characterisation from the previous lesson), so some term has $a_{N} > s - ε$ . Let $n \geq N$ . Then $s - ε < a_{N} \leq a_{n} \leq s$ , the middle step because the sequence increases and the last because $s$ is an upper bound. So $| a_{n} - s | < ε$ , and $a_{n} \to s$ . $□$

In $ℚ$ the theorem is false. The decimal truncations of $\sqrt{2}$ , namely $1, 1.4, 1.41, 1.414, 1.4142, \dots$ , are rational, increasing and bounded above by $2$ . In $ℝ$ the $n$ th truncation is within $10^{- (n - 1)}$ of $\sqrt{2}$ , and $10^{- (n - 1)} \leq \frac{1}{n}$ , so they converge to $\sqrt{2}$ , exactly as $\frac{1}{n}$ does. If they also converged to some rational $q$ (a real tolerance can always be shrunk to a rational one, so the definition means the same in either setting), uniqueness would force $q = \sqrt{2}$ , which is irrational. So the sequence has no limit in $ℚ$ . The proof breaks at its first line, where a supremum is needed and the rationals do not supply one. That is what completeness buys: sequences that ought to converge, do.

Writing a proof someone else can check

A proof is written for a reader who does not trust its writer, and four habits make it checkable. First, state what is being proved, with the quantifiers put back: "for every $ε > 0$ there is an $N$ " is a claim a reader can hold the proof against, while " $a_{n}$ gets close to $3$ " is not. Second, name every quantity when it is introduced and say where it comes from: "let $ε > 0$ " makes it arbitrary, "there is $N \in ℕ$ with $N > \frac{5}{ε}$ " makes it chosen, and a reader can then check that $N$ depends on $ε$ and on nothing chosen later. Third, justify each step by a definition or an earlier result, by name: the Archimedean property, the triangle inequality, the completeness axiom.

The fourth habit is to read the finished proof back as the sceptic, hunting for the move that would lose the game. Writing "take $N = \frac{5}{ε}$ " loses it, because that is not a natural number for most $ε$ , and the Archimedean property is what repairs it. Writing "let $ε = \frac{1}{n}$ " loses it too, because the sceptic chose $ε$ first and $n$ does not yet exist. A proof that survives that reading can be checked by anyone, line by line, without trusting its author.

What has been built

The course began with patterns that held for forty cases, or for every number below 906,150,257, and then failed, and it has built the means to settle such claims instead. Statements and quantifiers say exactly what is claimed and what its negation claims. Direct proof, cases, the contrapositive, contradiction and induction establish it. Sets, functions and relations are the language the claims are written in, bijections measure infinite sets, and completeness says what the real numbers are. This lesson needed nearly all of it: quantifiers for the definition, negation for divergence, contradiction for uniqueness, the supremum for monotone convergence.

Real Analysis continues directly from here, with the same $ε$ arguments applied to products and quotients of sequences, to continuity, derivatives and integrals, and to the question of which sequences converge when no limit is known in advance. Calculus uses these limits informally, as a tolerance game played with functions rather than sequences, and everything proved here stands behind it. Discrete Mathematics takes induction, sets, functions and relations as its daily tools. Each of them assumes a reader who can take a definition, say exactly what it claims, and prove things from it, which is what this course set out to make.