§7 Limits of Sequences

Why this matters

Derivatives, integrals, infinite sums and continuity are all defined through limits, so this one definition carries the weight of the whole subject. The goal of this section is to replace the vague “$s_n$ gets close to $s$” with a statement precise enough to prove things about.

Sequences

A sequence is a function whose domain is a set of integers of the form $\{n \in \mathbb{Z} : n \ge m\}$, where $m$ is usually $1$ or $0$. Instead of $s(n)$ we write $s_n$, and we write the whole sequence as $(s_n)_{n=m}^\infty$, as $(s_m, s_{m+1}, s_{m+2}, \dots)$, or just $(s_n)$. In this chapter every $s_n$ is a real number.

Example 1.

1. $s_n = \dfrac{1}{n^2}$ gives $\left(1, \tfrac14, \tfrac19, \tfrac1{16}, \dots\right)$.
2. $a_n = (-1)^n$ for $n \ge 0$ gives $(1, -1, 1, -1, \dots)$. Its set of values is just $\{-1, 1\}$, but the sequence itself has infinitely many terms.
3. $\cos\!\left(\tfrac{n\pi}{3}\right)$ cycles through $\tfrac12, -\tfrac12, -1, -\tfrac12, \tfrac12, 1, \dots$
4. $a_n = n^{1/n}$ gives $(1, 1.4142, 1.4422, 1.4142, 1.3797, \dots)$, with $a_{100} \approx 1.0471$ and $a_{1000} \approx 1.0069$.
5. $b_n = \left(1 + \tfrac1n\right)^n$ gives $(2, 2.25, 2.3704, 2.4414, \dots)$, with $b_{1000} \approx 2.7169$.

Keep the notation straight: parentheses $(\,\cdot\,)$ denote a sequence and braces $\{\,\cdot\,\}$ denote a set. Many results depend on which one you are working with.

The definition of a limit

Informally, $s$ is the limit of $(s_n)$ if $s_n$ is close to $s$ for all large $n$, not just for some of them. That is why $1$ is not a limit of $(-1)^n$: the even terms equal $1$, but the odd terms keep jumping back to $-1$.

7.1 Definition.

A sequence $(s_n)$ of real numbers converges to the real number $s$ if

$$\begin{gathered} \text{for each } \varepsilon > 0 \text{ there exists a number } N \text{ such that} \\ n > N \quad\text{implies}\quad |s_n - s| < \varepsilon. \end{gathered} \tag{1}$$

We then write $\lim_{n\to\infty} s_n = s$ or $s_n \to s$, and call $s$ the limit of $(s_n)$. A sequence that does not converge to any real number diverges.

Used in §8, §9, Theorem 9.1, §10

Some things to notice:

• By the Archimedean property, we may always take $N$ to be a positive integer.
• $\varepsilon$ is an ordinary positive real number. The interesting cases are the small ones.
• Condition (1) is infinitely many statements, one for each $\varepsilon > 0$. $N$ is allowed to depend on $\varepsilon$, and usually it must get larger as $\varepsilon$ gets smaller.

Intuition

Read $\varepsilon$ as a tolerance and $N$ as how long you must wait. Convergence says: name any tolerance, however strict, and from some point on every term stays within that tolerance of $s$, forever. Equivalently, every interval $(s - \varepsilon, s + \varepsilon)$ contains all but finitely many terms. One useful consequence: changing, inserting or deleting finitely many terms never affects whether a sequence converges, or what its limit is. You only need to take $N$ larger.

Common pitfall“closer and closer” is not the definition

The terms of $\frac1n$ get closer to $-1$ at every step, yet $\frac1n \to 0$, not $-1$. Shrinking distance is not enough: the distance has to drop below every $\varepsilon > 0$. Nor does the approach have to be steady. $\frac{(-1)^n}{n}$ hops from one side of $0$ to the other and still converges to $0$. Finally, $N$ may depend on $\varepsilon$ but never on $n$. “Take $N = n$” is not a valid choice.

The terms of $s_n = (-1)^n / n$ jump between positive and negative values forever. Does the sequence converge?

Yes, to $0$No, because the signs keep alternatingYes, to $1$

Try it: the ε–N game

Think of (1) as a game. Your opponent picks a band width $\varepsilon$. You must answer with an $N$ so that every point after $N$ lies inside the band $(s - \varepsilon,\, s + \varepsilon)$. The sequence converges to $s$ exactly when you can win no matter which $\varepsilon$ is chosen.

ε (band half-width)0.103Your N

Find the smallest N

n ≤ N (ignored)n > N, inside the bandn > N, outside the band

✗ N = 2 fails: n = 3 gives |s_n − 3/7| = 0.160 ≥ ε. The smallest N that works is 4.

Things to try:

• For $s_n = \frac{3n+1}{7n-4}$, make $\varepsilon$ smaller and watch the smallest working $N$ grow, roughly like $\frac{19}{49\varepsilon}$.
• For $(-1)^n$ with candidate limit $1$, no $N$ ever works once $\varepsilon \le 2$: there is always a later odd $n$ outside the band.
• Compare how fast $n^{1/n} \to 1$ and $(1 + 1/n)^n \to e$ by looking at how large $N$ must be for the same $\varepsilon$.

Try it: the adversarial ε game

The widget above lets you drag ε yourself. Here the roles are fixed, as in the definition: the page is an opponent that announces an $\varepsilon$, you commit to an $N$, and then it searches the whole tail $n > N$ for a point outside the band. Beat five ε in a row, down to $0.002$. (The same game for functions, with $\delta$ in place of $N$, is in §17.)

Then switch Who moves first? and play the swapped order, where you must name $N$ before seeing $\varepsilon$. The opponent now wins on every sequence except the constant one. This is exactly the difference between

$$\forall \varepsilon > 0\ \exists N\ \forall n > N:\ |s_n - s| < \varepsilon \qquad\text{and}\qquad \exists N\ \forall \varepsilon > 0\ \forall n > N:\ |s_n - s| < \varepsilon.$$

The second statement says $|s_n - s| < \varepsilon$ for every $\varepsilon$, i.e. $s_n = s$ for all $n > N$. Also try the “spiky” claim: five rounds can pass while the claim is false, because the opponent only needs a single far-out spike.

Round 1 of 5. The opponent announces ε = 0.500. Choose N so every n > N has |s_n − 3/7| < ε.

Your NSubmit N

n ≤ N (ignored)n > N, inside the bandn > N, outside the band

Example 2.

Let $s_n = \dfrac{3n+1}{7n-4}$. Writing $s_n = \dfrac{3 + 1/n}{7 - 4/n}$ suggests that $\lim s_n = \tfrac37$. By Definition 7.1, this claim means: for each $\varepsilon > 0$ there is an $N$ such that

$$n > N \implies \left| \frac{3n+1}{7n-4} - \frac37 \right| < \varepsilon.$$

In §8 we will show that $N = \dfrac{19}{49\varepsilon} + \dfrac47$ works. For $\varepsilon = 1, 0.1, 0.01, 0.001$ that gives $N \approx 0.96,\ 4.45,\ 39.35,\ 388.33$.

n	sn	|sn − 3/7|
2	0.7000	0.2714
3	0.5882	0.1597
4	0.5417	0.1131
5	0.5161	0.0876
6	0.5000	0.0714
40	0.4384	0.0098
400	0.4295	0.0010

A table of numbers can only check finitely many of these statements, so we will need proofs.

Example 3.

Back to Example 1:

1. $\lim \frac{1}{n^2} = 0$ (proved in §8, Example 1).
2. $(-1)^n$ diverges, so “$\lim a_n$” is meaningless here (§8, Example 4).
3. $\cos\!\left(\frac{n\pi}{3}\right)$ diverges (Exercise 8.7).
4. $\lim n^{1/n} = 1$ (Theorem 9.7(c)).
5. $\lim \left(1 + \frac1n\right)^n = e \approx 2.7182818$ (see §16 and §37).

Limits are unique

Theorem(Uniqueness of limits).

If $\lim s_n = s$ and $\lim s_n = t$, then $s = t$.

Proof

Let $\varepsilon > 0$. Since $s_n \to s$, there is $N_1$ such that $n > N_1$ implies $|s_n - s| < \frac{\varepsilon}{2}$. Since $s_n \to t$, there is $N_2$ such that $n > N_2$ implies $|s_n - t| < \frac{\varepsilon}{2}$. Take any $n > \max\{N_1, N_2\}$. By the Triangle Inequality,

$$|s - t| = |(s - s_n) + (s_n - t)| \le |s - s_n| + |s_n - t| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$

So $|s - t| < \varepsilon$ for every $\varepsilon > 0$, which forces $|s - t| = 0$, i.e. $s = t$.

∎

Lean 4 proof machine-checked

theorem lim_unique {s : ℕ → ℝ} {a b : ℝ} (ha : Lim s a) (hb : Lim s b) : a = b := by
  by_contra hne
  have hpos : 0 < |a - b| := abs_pos.mpr (sub_ne_zero.mpr hne)
  obtain ⟨N₁, h₁⟩ := ha (|a - b| / 2) (by linarith)
  obtain ⟨N₂, h₂⟩ := hb (|a - b| / 2) (by linarith)
  have h1 := h₁ (max N₁ N₂ + 1) (by omega)
  have h2 := h₂ (max N₁ N₂ + 1) (by omega)
  have key : |a - b| ≤ |s (max N₁ N₂ + 1) - a| + |s (max N₁ N₂ + 1) - b| := by
    calc |a - b| = |(s (max N₁ N₂ + 1) - b) - (s (max N₁ N₂ + 1) - a)| := by ring_nf
      _ ≤ |s (max N₁ N₂ + 1) - b| + |s (max N₁ N₂ + 1) - a| := abs_sub _ _
      _ = _ := add_comm _ _
  linarith

Checked against Mathlib in CI. Definitions: lean/Ross/Defs.lean.

Reading alongside Abbott

Abbott’s Section 2.2 uses the same definition. It also restates it in terms of neighbourhoods: $(s_n) \to s$ when every interval centred at $s$ contains all but finitely many terms. That version is worth keeping in mind; it is the one that generalizes to metric spaces (§13).

Practice: fill in the gaps

The scratch work is done; the formal proof runs forwards. The gaps are the algebra, the choice of $N$, and the inequality that connects them.

Fill in the gaps: lim (3n+1)/(n+2) = 3

This is the skeleton of the proof. In each gap choose the option that makes the argument work, then check.

1. Let $\varepsilon > 0$ and let $s_n = \frac{3n+1}{n+2}$. First, $|s_n - 3| = \left|\frac{3n + 1 - 3(n+2)}{n+2}\right| =$ $\frac{5}{n+2}$$\frac{7}{n+2}$$\frac{5}{n-2}$.
2. Choose $N =$ $\frac{5}{\varepsilon}$$\frac{\varepsilon}{5}$$5\varepsilon$. This depends only on $\varepsilon$, not on $n$.
3. Now let $n > N$. Then $|s_n - 3| = \frac{5}{n+2}$ $< \frac5n$$> \frac5n$$= \frac5n$ $< \varepsilon$, since $n > \frac5\varepsilon$.
4. So $|s_n - 3| < \varepsilon$ for every $n > N$, which is the definition of $\lim s_n = 3$.

Check your understanding

Take the quiz (3 questions)

When does $(s_n)$ fail to converge to $s$?

There is $\varepsilon > 0$ such that for every $N$, some $n > N$ has $|s_n - s| \ge \varepsilon$.For every $\varepsilon > 0$ and every $N$, all $n > N$ have $|s_n - s| \ge \varepsilon$.There is $\varepsilon > 0$ with $|s_n - s| \ge \varepsilon$ for every $n$.

In Definition 7.1, what does $N$ depend on?

On $\varepsilon$ (and on the sequence)On $n$Nothing: one $N$ must work for every $\varepsilon$.

Which sequence converges?

$s_n = (-1)^n$$s_n = n$$s_n = \frac{(-1)^n}{n}$

Exercises

Exercise 7.1

Write out the first five terms of the following sequences.

1. $s_n = \dfrac{1}{3n+1}$
2. $b_n = \dfrac{3n+1}{4n-1}$
3. $c_n = \dfrac{n}{3^n}$
4. $\sin\!\left(\dfrac{n\pi}{4}\right)$

Solution

1. $\tfrac14, \tfrac17, \tfrac1{10}, \tfrac1{13}, \tfrac1{16}$
2. $\tfrac43, 1, \tfrac{10}{11}, \tfrac{13}{15}, \tfrac{16}{19}$
3. $\tfrac13, \tfrac29, \tfrac19, \tfrac4{81}, \tfrac5{243}$
4. $\tfrac{\sqrt2}{2}, 1, \tfrac{\sqrt2}{2}, 0, -\tfrac{\sqrt2}{2}$

Tags: Warm-upComputation

Used in Exercise 7.2

Exercise 7.2

For each sequence in Exercise 7.1, determine whether it converges. If it converges, give its limit. No proofs are required.

Solution

1. converges to $0$
2. converges to $\tfrac34$
3. converges to $0$
4. diverges

For (d), the sequence cycles through $\tfrac{\sqrt2}{2}, 1, \tfrac{\sqrt2}{2}, 0, -\tfrac{\sqrt2}{2}, -1, -\tfrac{\sqrt2}{2}, 0, \dots$

Tags: Warm-upComputation

Exercise 7.3

For each sequence below, determine whether it converges and, if it converges, give its limit. No proofs are required.

1. $a_n = \dfrac{n}{n+1}$
2. $b_n = \dfrac{n^2+3}{n^2-3}$
3. $c_n = 2^{-n}$
4. $t_n = 1 + \dfrac2n$
5. $x_n = 73 + (-1)^n$
6. $s_n = 2^{1/n}$
7. $y_n = n!$
8. $d_n = (-1)^n n$
9. $\dfrac{(-1)^n}{n}$
10. $\dfrac{7n^3+8n}{2n^3-3}$
11. $\dfrac{9n^2-18}{6n+18}$
12. $\sin\dfrac{n\pi}{2}$
13. $\sin(n\pi)$
14. $\sin\dfrac{2n\pi}{3}$
15. $\dfrac1n \sin n$
16. $\dfrac{2^{n+1}+5}{2^n-7}$
17. $\dfrac{3^n}{n!}$
18. $\left(1+\dfrac1n\right)^2$
19. $\dfrac{4n^2+3}{3n^2-2}$
20. $\dfrac{6n+4}{9n^2+7}$

Hint

For rational functions of $n$, divide the numerator and denominator by the highest power of $n$ in the denominator. For (q), compare consecutive terms: $\frac{3^{n+1}/(n+1)!}{3^n/n!} = \frac{3}{n+1}$.

Solution

Converge: (a) $1$, (b) $1$, (c) $0$, (d) $1$, (f) $1$, (i) $0$, (j) $\tfrac72$, (m) $0$, (o) $0$, (p) $2$, (q) $0$, (r) $1$, (s) $\tfrac43$, (t) $0$.

Diverge: (e), (g), (h), (k), (l), (n).

Tags: Warm-upComputation

Exercise 7.4

Give examples of

1. a sequence $(x_n)$ of irrational numbers whose limit is a rational number;
2. a sequence $(r_n)$ of rational numbers whose limit is an irrational number.

Solution

(a) $x_n = \frac{\sqrt2}{n} \to 0$.

(b) Let $r_n$ be the decimal expansion of $\sqrt2$ cut off after $n$ digits: $1, 1.4, 1.41, 1.414, \dots \to \sqrt2$.

Tags: Warm-upExamples & counterexamples

Exercise 7.5

Determine the following limits. No proofs are required, but show any relevant algebra.

1. $\lim s_n$ where $s_n = \sqrt{n^2+1} - n$
2. $\lim\left(\sqrt{n^2+n} - n\right)$
3. $\lim\left(\sqrt{4n^2+n} - 2n\right)$

Hint

For (a), first show $s_n = \dfrac{1}{\sqrt{n^2+1}+n}$ by multiplying and dividing by the conjugate $\sqrt{n^2+1}+n$.

Solution

(a) $s_n = \dfrac{1}{\sqrt{n^2+1}+n} \to 0$.

(b) $\sqrt{n^2+n}-n = \dfrac{n}{\sqrt{n^2+n}+n} = \dfrac{1}{\sqrt{1+1/n}+1} \to \dfrac12$.

(c) $\sqrt{4n^2+n}-2n = \dfrac{n}{\sqrt{4n^2+n}+2n} = \dfrac{1}{\sqrt{4+1/n}+2} \to \dfrac14$.

Tags: StandardComputation

Used in Exercise 8.8