§9 Limit Theorems for Sequences

This section proves the familiar rules for computing limits. First: a sequence $(s_n)$ is bounded if $\{s_n : n \in \mathbb{N}\}$ is a bounded set, i.e. there is a constant $M$ with $|s_n| \le M$ for all $n$.

Why this matters

Proving every limit straight from Definition 7.1 would be exhausting, as Example 3 of §8 showed. The theorems below do the ε–N work once, for sums, products and quotients, so that afterwards most limits can be computed by algebra. Each proof is a model of a technique you will reuse: split the error with the Triangle Inequality, and control each piece separately.

Algebra of limits

9.1 Theorem.

Convergent sequences are bounded.

Used in Exercise 8.7, Theorem 9.4, Lemma 10.10, §18

Proof

Let $s = \lim s_n$. Applying Definition 7.1 with $\varepsilon = 1$, there is $N \in \mathbb{N}$ with $|s_n - s| < 1$ for $n > N$, so $|s_n| < |s| + 1$ for $n > N$ by the Triangle Inequality. Let $M = \max\{|s| + 1, |s_1|, |s_2|, \dots, |s_N|\}$. Then $|s_n| \le M$ for all $n$.

∎

The proof used the definition of limit for just one value of $\varepsilon$, and the choice $\varepsilon = 1$ was arbitrary.

9.2 Theorem.

If $(s_n)$ converges to $s$ and $k \in \mathbb{R}$, then $(ks_n)$ converges to $ks$. That is, $\lim(ks_n) = k \cdot \lim s_n$.

Used in Exercise 9.1, Exercise 9.3, Exercise 10.2, Exercise 11.8, §17, Theorem 17.3

Proof

The case $k = 0$ is trivial, so assume $k \ne 0$. Let $\varepsilon > 0$. There is $N$ with $|s_n - s| < \frac{\varepsilon}{|k|}$ for $n > N$. Then $n > N$ implies $|ks_n - ks| < \varepsilon$.

∎

9.3 Theorem.

If $(s_n)$ converges to $s$ and $(t_n)$ converges to $t$, then $(s_n + t_n)$ converges to $s + t$. That is, $\lim(s_n + t_n) = \lim s_n + \lim t_n$.

Used in Theorem 9.7, Exercise 9.1, Exercise 9.3, Exercise 12.4, Theorem 17.4

Proof

Let $\varepsilon > 0$. Note $|s_n + t_n - (s + t)| \le |s_n - s| + |t_n - t|$. Choose $N_1$ with $|s_n - s| < \frac\varepsilon2$ for $n > N_1$ and $N_2$ with $|t_n - t| < \frac\varepsilon2$ for $n > N_2$. With $N = \max\{N_1, N_2\}$, $n > N$ implies $|s_n + t_n - (s + t)| < \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon$.

∎

9.4 Theorem.

If $(s_n)$ converges to $s$ and $(t_n)$ converges to $t$, then $(s_n t_n)$ converges to $st$. That is, $\lim(s_n t_n) = (\lim s_n)(\lim t_n)$.

Used in Theorem 9.6, Exercise 9.1, Exercise 9.3, Exercise 12.8, §17, Theorem 17.4

Discussion.

The trick is to add and subtract $s_n t$:

$$|s_n t_n - st| = |s_n t_n - s_n t + s_n t - st| \le |s_n|\cdot|t_n - t| + |t|\cdot|s_n - s|.$$

For large $n$, $|t_n - t|$ and $|s_n - s|$ are small, and $|t|$ is a constant. Theorem 9.1 says $|s_n|$ is bounded, so the first term can be made small too.

Proof

Let $\varepsilon > 0$. By Theorem 9.1 there is $M > 0$ with $|s_n| \le M$ for all $n$. Choose $N_1$ with $|t_n - t| < \frac{\varepsilon}{2M}$ for $n > N_1$, and $N_2$ with $|s_n - s| < \frac{\varepsilon}{2(|t| + 1)}$ for $n > N_2$. (We use $2(|t| + 1)$ rather than $2|t|$ because $t$ could be $0$.) For $n > \max\{N_1, N_2\}$,

$$|s_n t_n - st| \le M \cdot \frac{\varepsilon}{2M} + |t| \cdot \frac{\varepsilon}{2(|t|+1)} < \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon.$$∎

Intuitionwhere the product error comes from

Write $s_n = s + e_n$ and $t_n = t + f_n$, where the errors $e_n$ and $f_n$ tend to $0$. Then

$$s_n t_n - st = s f_n + t e_n + e_n f_n.$$

Each piece is a fixed number times something small, or small times small, so the whole error is small. The proof above organizes the same idea differently: it uses the bound $|s_n| \le M$ from Theorem 9.1 so that only two pieces are needed.

Suppose $s_n \to 3$ and $t_n \to -2$. What happens to the product $s_n t_n$?

It converges to $-6$It converges, but the limit depends on how the two sequences approach their limitsIt need not converge

The widget below runs these two proofs with numbers. For a given $\varepsilon$ it finds the $N_1$ and $N_2$ the proof asks for and takes $N = \max\{N_1, N_2\}$. It also shows the smallest $N$ that actually works, which is often much smaller: a proof only needs some $N$.

ε0.320

sₙ + tₙ inside the ε-bandoutside the bandN₁ (dashed)N₂ (dashed)smallest N that works

The proof asks for |sₙ − 2| < ε/2 = 0.160, which holds for n > N₁ = 39, and |tₙ − 0| < ε/2, which holds for n > N₂ = 6.

✓ So N = max{39, 6} = 39 works: every n > 39 has |sₙ + tₙ − (2 + 0)| < 0.320.

The smallest N that actually works is 15. The proof's N is safe but not sharp: the triangle inequality assumes the two errors pile up, when often they partly cancel.

To handle quotients we first deal with reciprocals.

9.5 Lemma.

If $(s_n)$ converges to $s$, $s_n \ne 0$ for all $n$, and $s \ne 0$, then $(1/s_n)$ converges to $1/s$.

Used in Theorem 9.6, Theorem 9.7, Theorem 12.1

Discussion.

$\frac1{s_n} - \frac1s = \frac{s - s_n}{s_n s}$. The numerator is small for large $n$. The only danger is a small denominator, and Example 6 of §8 rules that out: $m = \inf\{|s_n| : n \in \mathbb{N}\} > 0$. So $\left|\frac1{s_n} - \frac1s\right| \le \frac{|s - s_n|}{m|s|}$.

Proof

Let $\varepsilon > 0$. By Example 6 of §8 there is $m > 0$ with $|s_n| \ge m$ for all $n$. Choose $N$ with $|s - s_n| < \varepsilon \cdot m|s|$ for $n > N$. Then $n > N$ implies

$$\left| \frac1{s_n} - \frac1s \right| = \frac{|s - s_n|}{|s_n s|} \le \frac{|s - s_n|}{m|s|} < \varepsilon.$$∎

9.6 Theorem.

Suppose $(s_n)$ converges to $s$ and $(t_n)$ converges to $t$. If $s \ne 0$ and $s_n \ne 0$ for all $n$, then $(t_n/s_n)$ converges to $t/s$.

Used in Exercise 9.1, Exercise 9.3, Theorem 17.4

Proof

By Lemma 9.5, $(1/s_n)$ converges to $1/s$, so by Theorem 9.4, $\lim \frac{t_n}{s_n} = \lim \frac1{s_n} \cdot t_n = \frac1s \cdot t = \frac ts$.

∎

Common pitfallcheck the hypotheses first

• The theorems need both limits to exist as real numbers. With $s_n = (-1)^n$ and $t_n = -(-1)^n$, the sum $s_n + t_n = 0$ converges although neither sequence does. So “$\lim(s_n + t_n) = \lim s_n + \lim t_n$” can only be written once you know the right side makes sense.
• Theorem 9.6 needs $\lim s_n \ne 0$. When the denominator tends to $0$, rewrite the quotient first, as in Examples 2 and 3 below.
• Limits preserve $\le$ but not $<$. Every term of $\frac1n$ is positive, yet $\lim \frac1n = 0$. Exercise 8.9 and Exercise 9.9(c) give the correct statements.

These theorems, together with a few standard examples, make many limits easy.

9.7 Theorem (Basic Examples).

1. $\lim_{n\to\infty} \dfrac1{n^p} = 0$ for $p > 0$.
2. $\lim_{n\to\infty} a^n = 0$ if $|a| < 1$.
3. $\lim n^{1/n} = 1$.
4. $\lim_{n\to\infty} a^{1/n} = 1$ for $a > 0$.

Used in §1, §7, Exercise 9.1, Exercise 9.7, Exercise 9.12, Exercise 9.13, Exercise 9.14, Theorem 12.2, Exercise 12.11, §14, Theorem 14.9

Proof

(a) Let $\varepsilon > 0$ and $N = (1/\varepsilon)^{1/p}$. Then $n > N$ implies $n^p > \frac1\varepsilon$, so $0 < \frac1{n^p} < \varepsilon$. (The meaning of $n^p$ for non-integer $p$ is discussed in §37.)

(b) We may assume $a \ne 0$. Since $|a| < 1$, write $|a| = \frac1{1+b}$ with $b > 0$. By the binomial theorem (Exercise 1.12), $(1 + b)^n \ge 1 + nb > nb$, so $|a^n - 0| = \frac1{(1+b)^n} < \frac1{nb}$. Given $\varepsilon > 0$, let $N = \frac1{\varepsilon b}$. Then $n > N$ implies $|a^n - 0| < \frac1{nb} < \varepsilon$.

(c) Let $s_n = n^{1/n} - 1 \ge 0$. By Theorem 9.3 it suffices to show $\lim s_n = 0$. Since $n = (1 + s_n)^n$, the binomial expansion gives, for $n \ge 2$,

$$n = (1 + s_n)^n \ge 1 + ns_n + \tfrac12 n(n-1)s_n^2 > \tfrac12 n(n-1)s_n^2.$$

So $s_n^2 < \frac2{n-1}$, i.e. $s_n < \sqrt{\frac2{n-1}}$ for $n \ge 2$. A standard argument now gives $\lim s_n = 0$ (Exercise 9.7).

(d) If $a \ge 1$, then for $n \ge a$ we have $1 \le a^{1/n} \le n^{1/n}$, and (c) with the squeeze lemma (Exercise 8.5(a)) gives $\lim a^{1/n} = 1$. If $0 < a < 1$, then $\frac1a > 1$, so $\lim (1/a)^{1/n} = 1$, and Lemma 9.5 gives $\lim a^{1/n} = 1$.

∎

Example 1.

Prove $\lim s_n = \frac14$, where $s_n = \dfrac{n^3 + 6n^2 + 7}{4n^3 + 3n - 4}$.

Solution

Divide by $n^3$: $s_n = \dfrac{1 + \frac6n + \frac7{n^3}}{4 + \frac3{n^2} - \frac4{n^3}}$. By Theorem 9.7(a), $\lim\frac1n = \lim\frac1{n^2} = \lim\frac1{n^3} = 0$. By Theorems 9.3 and 9.2, $\lim\left(1 + \frac6n + \frac7{n^3}\right) = 1$ and $\lim\left(4 + \frac3{n^2} - \frac4{n^3}\right) = 4$. Theorem 9.6 gives $\lim s_n = \frac14$.

∎

Example 2.

Find $\lim \dfrac{n - 5}{n^2 + 7}$.

Solution

Writing it as $\frac{1 - 5/n}{n + 7/n}$ is no good, because the denominator does not converge. Instead divide by $n^2$: $\dfrac{\frac1n - \frac5{n^2}}{1 + \frac7{n^2}}$. The numerator tends to $0$ and the denominator to $1$, so the limit is $\frac01 = 0$.

∎

Example 3.

Find $\lim \dfrac{n^2 + 3}{n + 1}$.

Solution

Both $\dfrac{n + \frac3n}{1 + \frac1n}$ and $\dfrac{1 + \frac3{n^2}}{\frac1n + \frac1{n^2}}$ lead to trouble: either the numerator diverges or the denominator tends to $0$. In fact the sequence does not converge, and $\lim \frac{n^2+3}{n+1}$ is undefined for now. You may feel the urge to write $+\infty$. Definition 9.8 makes that legitimate; see Example 6.

∎

Infinite limits

9.8 Definition.

For a sequence $(s_n)$, we write $\lim s_n = +\infty$ provided

$$\text{for each } M > 0 \text{ there is a number } N \text{ such that } n > N \text{ implies } s_n > M.$$

We then say $(s_n)$ diverges to $+\infty$. Similarly, $\lim s_n = -\infty$ provided for each $M < 0$ there is $N$ such that $n > N$ implies $s_n < M$.

Used in Exercise 9.17

From now on, $(s_n)$ has a limit, or the limit exists, if $(s_n)$ converges, diverges to $+\infty$, or diverges to $-\infty$. For $+\infty$ the challenging values of $M$ are large positive numbers; for $-\infty$ they are “large” negative numbers like $-10{,}000{,}000{,}000$.

Example 4.

$\lim n^2 = +\infty$, $\lim(-n) = -\infty$, $\lim 2^n = +\infty$ and $\lim(\sqrt n + 7) = +\infty$. On the other hand $s_n = (-1)^n n$ and $t_n = n\cos^2\left(\frac{n\pi}2\right)$ are unbounded but diverge neither to $+\infty$ nor to $-\infty$, so $\lim[(-1)^n n]$ and $\lim[n\cos^2(\frac{n\pi}2)]$ are meaningless. Note $t_n = n$ for even $n$ and $t_n = 0$ for odd $n$.

Proofs about infinite limits follow the same strategy as for finite limits.

Example 5.

Give a formal proof that $\lim(\sqrt n + 7) = +\infty$.

Discussion.

Given $M > 0$ we want $\sqrt n + 7 > M$, which holds when $\sqrt n > M - 7$, i.e. when $n > (M - 7)^2$. So take $N = (M - 7)^2$.

Formal Proof

Let $M > 0$ and $N = (M - 7)^2$. Then $n > N$ implies $\sqrt n > M - 7$, hence $\sqrt n + 7 > M$.

∎

Example 6.

Give a formal proof that $\lim \dfrac{n^2+3}{n+1} = +\infty$ (see Example 3).

Discussion.

Bound the fraction below by a multiple of $\frac{n^2}{n} = n$. Since $n^2 + 3 > n^2$ and $n + 1 \le 2n$, we have $\frac{n^2+3}{n+1} > \frac{n^2}{2n} = \frac n2$, so it suffices to make $\frac n2 > M$.

Formal Proof

Let $M > 0$ and $N = 2M$. Then $n > N$ implies $\dfrac{n^2+3}{n+1} > \dfrac{n^2}{2n} = \dfrac n2 > M$.

∎

Warning

Do not apply the limit Theorems 9.2–9.6 to infinite limits. Use Theorem 9.9 or 9.10 below, or Exercises 9.9–9.12.

Common pitfallindeterminate forms

Knowing that one factor tends to $+\infty$ and the other to $0$ tells you nothing by itself: $n \cdot \frac1n \to 1$, $n^2 \cdot \frac1n \to +\infty$ and $n \cdot \frac1{n^2} \to 0$. Theorems 9.9 and 9.10 are the safe statements; they avoid these cases by requiring $\lim t_n > 0$, or by working with reciprocals of positive terms.

9.9 Theorem.

Let $(s_n)$ and $(t_n)$ be sequences with $\lim s_n = +\infty$ and $\lim t_n > 0$ ($\lim t_n$ may be finite or $+\infty$). Then $\lim s_n t_n = +\infty$.

Used in Exercise 9.16, Theorem 12.1

Discussion.

Given $M > 0$ we need $s_n t_n > M$ for large $n$. We must keep the $t_n$ away from $0$: pick $m$ with $0 < m < \lim t_n$, so that $t_n > m$ for large $n$. Then it is enough to have $s_n > \frac Mm$.

Proof

Let $M > 0$ and choose a real $m$ with $0 < m < \lim t_n$. Whether or not $\lim t_n = +\infty$, there is $N_1$ such that $n > N_1$ implies $t_n > m$ (Exercise 8.10). Since $\lim s_n = +\infty$, there is $N_2$ such that $n > N_2$ implies $s_n > \frac Mm$. For $n > \max\{N_1, N_2\}$, $s_n t_n > \frac Mm \cdot m = M$.

∎

Example 7.

Use Theorem 9.9 to prove $\lim \dfrac{n^2+3}{n+1} = +\infty$.

Solution

$\dfrac{n^2+3}{n+1} = \dfrac{n + \frac3n}{1 + \frac1n} = s_n t_n$ with $s_n = n + \frac3n$ and $t_n = \dfrac1{1 + \frac1n}$. Clearly $\lim s_n = +\infty$ and $\lim t_n = 1$, so Theorem 9.9 applies.

∎

9.10 Theorem.

For a sequence $(s_n)$ of positive real numbers, $\lim s_n = +\infty$ if and only if $\lim \left(\dfrac1{s_n}\right) = 0$.

Used in Exercise 9.12, Exercise 9.13, Exercise 9.16

Proof

We must show

$$\lim s_n = +\infty \implies \lim\frac1{s_n} = 0 \tag{1}$$$$\lim\frac1{s_n} = 0 \implies \lim s_n = +\infty. \tag{2}$$

The two proofs look alike, but the thought processes are quite different.

(1) Suppose $\lim s_n = +\infty$. Let $\varepsilon > 0$ and $M = \frac1\varepsilon$. There is $N$ with $s_n > M = \frac1\varepsilon$ for $n > N$. Then $n > N$ implies $\varepsilon > \frac1{s_n} > 0$, so $\left|\frac1{s_n} - 0\right| < \varepsilon$.

(2) Starting afresh, suppose $\lim \frac1{s_n} = 0$. Let $M > 0$ and $\varepsilon = \frac1M$. There is $N$ with $\left|\frac1{s_n} - 0\right| < \frac1M$ for $n > N$. Since $s_n > 0$, this says $0 < \frac1{s_n} < \frac1M$, so $s_n > M$.

∎

Reading alongside Abbott

Abbott’s Section 2.3 proves the same rules as one “Algebraic Limit Theorem”, and adds an “Order Limit Theorem” that collects the facts about inequalities found here in Exercises 8.9 and 9.9.

Practice: put the proof in order

The idea behind Theorem 9.1: the tail is trapped near the limit, and a finite list has a largest member.

Put the proof in order: Convergent sequences are bounded (Theorem 9.1)

These are the steps of the proof, shuffled. Use the buttons to move each step into place, then check.

1. Let $(s_n)$ converge to $s$. Apply the definition of convergence with $\varepsilon = 1$: there is $N$ such that $|s_n - s| < 1$ for all $n > N$.
2. For $n > N$ the triangle inequality gives $|s_n| \le |s| + |s_n - s| < |s| + 1$.
3. That controls every term after the $N$th; only the finitely many terms $s_1, \dots, s_N$ are left over.
4. Set $M = \max\{|s| + 1,\ |s_1|,\ \dots,\ |s_N|\}$.
5. Then $|s_n| \le M$ for every $n$, so $(s_n)$ is bounded.

Practice: fill in the gaps

The proof of the sum rule (Theorem 9.3), with the key choices left blank.

Fill in the gaps: Limit of a sum (Theorem 9.3)

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

1. Let $s_n \to s$ and $t_n \to t$, and let $\varepsilon > 0$.
2. Since $s_n \to s$, there is $N_1$ with $|s_n - s| <$ $\frac\varepsilon2$$\varepsilon$$2\varepsilon$ for $n > N_1$. In the same way there is $N_2$ with $|t_n - t| <$ the same bound for $n > N_2$.
3. Let $N =$ $\max\{N_1, N_2\}$$\min\{N_1, N_2\}$either one of them. Then for $n > N$ both estimates hold.
4. For $n > N$, $|s_n - s| + |t_n - t|$$|s_n - s| - |t_n - t|$$|s_n - s| \cdot |t_n - t|$ is an upper bound for $|(s_n + t_n) - (s + t)|$, and it is less than $\frac\varepsilon2 + \frac\varepsilon2 = \varepsilon$.

Practice: spot the error

Limit laws come with hypotheses. Find the line that ignores one.

Spot the error: a limit with a square root

One line of this argument is wrong. Press the button on the line you think is the mistake.

1. Claim: $\lim\left(\sqrt{n^2 + n} - n\right) = 0$.
2. Both $\sqrt{n^2 + n}$ and $n$ tend to $+\infty$.
3. So $\lim\left(\sqrt{n^2 + n} - n\right) = \lim\sqrt{n^2 + n} - \lim n = \infty - \infty = 0$.
4. Hence the limit is $0$.

Now write a proof yourself. Each line you type is checked against a few allowed moves.

Write the proof: a sum of convergent sequences

Check your understanding

Take the quiz (3 questions)

Suppose $(s_n)$ is bounded. Must $(s_n)$ converge?

Yes, by Theorem 9.1.No. For example, $s_n = (-1)^n$.

What is $\lim \frac{3n^2 + 1}{n^2 + 5n}$?

$3$$0$$+\infty$

Suppose $\lim s_n = +\infty$ and $\lim t_n = 0$. What is $\lim s_n t_n$?

It can be anything; Theorem 9.9 does not apply.$0$$+\infty$

Exercises

Exercise 9.1

Using the limit Theorems 9.2–9.7, prove the following. Justify all steps.

1. $\lim \dfrac{n+1}{n} = 1$
2. $\lim \dfrac{3n+7}{6n-5} = \dfrac12$
3. $\lim \dfrac{17n^5 + 73n^4 - 18n^2 + 3}{23n^5 + 13n^3} = \dfrac{17}{23}$

Hint

Divide numerator and denominator by the highest power of $n$ in the denominator, then use 9.7(a), 9.2, 9.3 and 9.6.

Solution

We use $\lim \frac1n = 0$ (Theorem 9.7(a)), and note that $\frac1{n^k} = \left(\frac1n\right)^k \to 0$ for each $k \in \mathbb{N}$ by repeated use of Theorem 9.4.

1. $\dfrac{n+1}{n} = 1 + \dfrac1n \to 1 + 0 = 1$ by Theorem 9.3.
2. Dividing top and bottom by $n$, $\dfrac{3n+7}{6n-5} = \dfrac{3 + 7\cdot\frac1n}{6 - 5\cdot\frac1n}$. By Theorems 9.2 and 9.3 the numerator tends to $3$ and the denominator to $6 \neq 0$, so Theorem 9.6 gives the limit $\frac36 = \frac12$.
3. Dividing top and bottom by $n^5$, the fraction is $\frac{17 + 73\cdot\frac1n - 18\cdot\frac1{n^3} + 3\cdot\frac1{n^5}}{23 + 13\cdot\frac1{n^2}}.$ The numerator tends to $17$, the denominator to $23 \ne 0$ (Theorems 9.2, 9.3 and the remark above), so the quotient tends to $\frac{17}{23}$ by Theorem 9.6.

Tags: Warm-upLimit laws

Exercise 9.2

Suppose $\lim x_n = 3$, $\lim y_n = 7$ and all $y_n$ are nonzero. Determine:

1. $\lim(x_n + y_n)$
2. $\lim \dfrac{3y_n - x_n}{y_n^2}$

Solution

(a) $10$ (b) $\dfrac{21 - 3}{49} = \dfrac{18}{49}$

Tags: Warm-upLimit laws

Exercise 9.3

Suppose $\lim a_n = a$, $\lim b_n = b$, and $s_n = \dfrac{a_n^3 + 4a_n}{b_n^2 + 1}$. Prove $\lim s_n = \dfrac{a^3 + 4a}{b^2 + 1}$ carefully, using the limit theorems.

Hint

$a_n^3 = a_n \cdot a_n \cdot a_n$ (Theorem 9.4 twice). The denominators $b_n^2 + 1$ are never $0$ and tend to $b^2 + 1 \ne 0$.

Solution

By Theorem 9.4 (applied twice), $a_n^3 = a_n \cdot a_n \cdot a_n \to a^3$, and $4a_n \to 4a$ (Theorem 9.2), so by Theorem 9.3 the numerator tends to $a^3 + 4a$. Similarly $b_n^2 \to b^2$, so $b_n^2 + 1 \to b^2 + 1$. Each $b_n^2 + 1 \ge 1$ is nonzero and the limit $b^2 + 1 \ge 1$ is nonzero, so Theorem 9.6 applies and $\lim s_n = \frac{a^3 + 4a}{b^2 + 1}.$

Tags: StandardLimit laws

Exercise 9.4

Let $s_1 = 1$ and $s_{n+1} = \sqrt{s_n + 1}$ for $n \ge 1$.

1. List the first four terms of $(s_n)$.
2. It turns out that $(s_n)$ converges. Assume this and prove the limit is $\frac12(1 + \sqrt5)$.

Solution

(a) $1, \sqrt2, \sqrt{\sqrt2 + 1}, \sqrt{\sqrt{\sqrt2+1}+1}$, approximately $1, 1.4142, 1.5538, 1.5981$. (b) If $s = \lim s_n$, then $\lim s_{n+1} = s$ too, and by Example 5 of §8, $s = \sqrt{s + 1}$. So $s^2 - s - 1 = 0$ and $s = \frac{1 \pm \sqrt5}{2}$. Since $s_n \ge 1$, $s \ge 1$ (Exercise 8.9), so $s = \frac{1 + \sqrt5}{2}$.

Tags: StandardLimit lawsComputation

Exercise 9.5

Let $t_1 = 1$ and $t_{n+1} = \dfrac{t_n^2 + 2}{2t_n}$ for $n \ge 1$. Assume $(t_n)$ converges and find the limit.

Solution

The limit $t$ satisfies $t = \frac{t^2+2}{2t}$, so $t^2 = 2$. All $t_n > 0$, so $t = \sqrt2$. (This is Newton’s method for $x^2 - 2$; compare §10.)

Tags: Warm-upLimit lawsComputation

Exercise 9.6

Let $x_1 = 1$ and $x_{n+1} = 3x_n^2$ for $n \ge 1$.

1. Show that if $a = \lim x_n$, then $a = \frac13$ or $a = 0$.
2. Does $\lim x_n$ exist? Explain.
3. Discuss the apparent contradiction between parts (a) and (b).

Solution

(a) $a = 3a^2$. (b) $x_n = 3^{2^{n-1} - 1}$, so $\lim x_n = +\infty$. (c) Part (a) only shows if the sequence converges, its limit must be $0$ or $\frac13$. It does not show the sequence converges, and it doesn’t.

Tags: StandardLimit lawsExplain why

Exercise 9.7

Complete the proof of Theorem 9.7(c), i.e. give the standard argument needed to show $\lim s_n = 0$.

Hint

Given $\varepsilon > 0$, $\sqrt{\frac{2}{n-1}} < \varepsilon$ when $n > 1 + \frac2{\varepsilon^2}$. Take $N = 1 + \frac{2}{\varepsilon^2}$.

Solution

Let $\varepsilon > 0$ and put $N = 1 + \frac{2}{\varepsilon^2}$. If $n > N$ then $n - 1 > \frac{2}{\varepsilon^2}$, so $\frac{2}{n-1} < \varepsilon^2$ and $\sqrt{\frac{2}{n-1}} < \varepsilon$. In the proof of (c) we have $0 \le s_n \le \sqrt{\frac{2}{n-1}}$, so $|s_n - 0| < \varepsilon$ for all $n > N$. Hence $\lim s_n = 0$.

Tags: Standardε–N / ε–δ

Used in Theorem 9.7

Exercise 9.8

Give the following when they exist. Otherwise assert “NOT EXIST.”

1. $\lim n^3$
2. $\lim(-n^3)$
3. $\lim(-n)^n$
4. $\lim(1.01)^n$
5. $\lim n^n$

Solution

1. $+\infty$
2. $-\infty$
3. NOT EXIST
4. $+\infty$
5. $+\infty$

Tags: Warm-upComputation

Exercise 9.9

Suppose there is $N_0$ such that $s_n \le t_n$ for all $n > N_0$.

1. Prove that if $\lim s_n = +\infty$, then $\lim t_n = +\infty$.
2. Prove that if $\lim t_n = -\infty$, then $\lim s_n = -\infty$.
3. Prove that if $\lim s_n$ and $\lim t_n$ exist, then $\lim s_n \le \lim t_n$.

Hint

For (c), if both limits are finite, apply Exercise 8.9 to $t_n - s_n \ge 0$. The infinite cases follow from (a) and (b).

Solution

1. Let $M \in \mathbb{R}$. Choose $N_1$ with $s_n > M$ for $n > N_1$. For $n > \max\{N_0, N_1\}$ we get $t_n \ge s_n > M$. So $t_n \to +\infty$.
2. Let $M \in \mathbb{R}$. Choose $N_1$ with $t_n < M$ for $n > N_1$. For $n > \max\{N_0, N_1\}$, $s_n \le t_n < M$. So $s_n \to -\infty$.
3. If $\lim s_n = +\infty$, then $\lim t_n = +\infty$ by (a), so the inequality holds; likewise it holds if $\lim t_n = -\infty$ by (b), and trivially if $\lim s_n = -\infty$ or $\lim t_n = +\infty$. The remaining case is that both limits are finite, say $s_n \to s$ and $t_n \to t$. Suppose $t < s$ and let $\varepsilon = \frac{s - t}{2}$. For large $n$ we have $s_n > s - \varepsilon$ and $t_n < t + \varepsilon = s - \varepsilon$, so $t_n < s_n$, contradicting $s_n \le t_n$. Hence $s \le t$.

Tags: Standardε–N / ε–δ

Used in Exercise 9.16, Exercise 12.1, Exercise 12.4, Exercise 12.8

Exercise 9.10

1. Show that if $\lim s_n = +\infty$ and $k > 0$, then $\lim(ks_n) = +\infty$.
2. Show $\lim s_n = +\infty$ if and only if $\lim(-s_n) = -\infty$.
3. Show that if $\lim s_n = +\infty$ and $k < 0$, then $\lim(ks_n) = -\infty$.

Solution

1. Let $M > 0$. Since $s_n \to +\infty$, there is $N$ with $s_n > M/k$ for $n > N$. Then $ks_n > M$ for $n > N$.
2. $\lim s_n = +\infty$ means: for every $M$ there is $N$ with $s_n > M$ for $n > N$. Replacing $M$ by $-M$ (which ranges over all reals as $M$ does), this says: for every $M$ there is $N$ with $-s_n < M$ for $n > N$, which is $\lim(-s_n) = -\infty$.
3. Since $-k > 0$, part (a) gives $(-k)s_n \to +\infty$, and then part (b) gives $ks_n = -\bigl((-k)s_n\bigr) \to -\infty$.

Tags: Standardε–N / ε–δ

Used in Exercise 10.5, Exercise 11.8

Exercise 9.11

1. Show that if $\lim s_n = +\infty$ and $\inf\{t_n : n \in \mathbb{N}\} > -\infty$, then $\lim(s_n + t_n) = +\infty$.
2. Show that if $\lim s_n = +\infty$ and $\lim t_n > -\infty$, then $\lim(s_n + t_n) = +\infty$.
3. Show that if $\lim s_n = +\infty$ and $(t_n)$ is bounded, then $\lim(s_n + t_n) = +\infty$.

Hint

(a) Let $m = \inf t_n$. Given $M$, make $s_n > M - m$. (b) If $\lim t_n > -\infty$, then $(t_n)$ is bounded below (use Exercise 8.10 for the tail and handle the finitely many early terms separately).

Solution

1. Let $m = \inf\{t_n\}$, a real number. Given $M$, choose $N$ with $s_n > M - m$ for $n > N$. Then $s_n + t_n \ge s_n + m > M$.
2. If $\lim t_n > -\infty$, pick a real $c < \lim t_n$. Then $t_n > c$ for $n > N_1$ (Exercise 8.10), and the finitely many terms $t_1, \dots, t_{N_1}$ have a minimum, so $\inf\{t_n\} \ge \min\{c, t_1, \dots, t_{N_1}\} > -\infty$. Now apply (a).
3. A bounded sequence has $\inf\{t_n\} > -\infty$, so (a) applies.

Tags: Standardε–N / ε–δ

Used in Exercise 9.16

Exercise 9.12

Assume all $s_n \ne 0$ and that the limit $L = \lim \left|\dfrac{s_{n+1}}{s_n}\right|$ exists. (This exercise is referred to in several places.)

1. Show that if $L < 1$, then $\lim s_n = 0$.
2. Show that if $L > 1$, then $\lim |s_n| = +\infty$.

Hint

(a) Select $a$ with $L < a < 1$ and $N$ with $|s_{n+1}| < a|s_n|$ for $n \ge N$. Then $|s_n| < a^{n-N}|s_N|$ for $n > N$. (b) Apply (a) to $t_n = \frac1{|s_n|}$ and use Theorem 9.10.

Solution

1. Since $L \ge 0$, pick $a$ with $L < a < 1$ (so $0 < a < 1$). Because $\left|\frac{s_{n+1}}{s_n}\right| \to L < a$, there is $N$ with $|s_{n+1}| < a|s_n|$ for $n \ge N$. By induction, $|s_n| < a^{n-N}|s_N|$ for $n > N$, that is, $|s_n| < C a^n$ with $C = |s_N|a^{-N}$. Since $a^n \to 0$ (Theorem 9.7(b)), $|s_n| \to 0$ by squeezing, and $s_n \to 0$.
2. Let $t_n = \frac1{|s_n|}$. Then $\left|\frac{t_{n+1}}{t_n}\right| = \left|\frac{s_n}{s_{n+1}}\right| \to \frac1L < 1$ (including the case $L = +\infty$, where the limit is $0$). By (a), $t_n \to 0$, and since $t_n > 0$, Theorem 9.10 gives $|s_n| \to +\infty$.

Tags: Challengeε–N / ε–δInequalities

Used in Exercise 9.14, Exercise 9.15

Exercise 9.13

Show

$$\lim_{n\to\infty} a^n = \begin{cases} 0 & \text{if } |a| < 1 \\ 1 & \text{if } a = 1 \\ +\infty & \text{if } a > 1 \\ \text{does not exist} & \text{if } a \le -1. \end{cases}$$

Solution

$|a| < 1$: Theorem 9.7(b). $a = 1$: the sequence is constantly $1$.

$a > 1$: $(1/a)^n \to 0$ by Theorem 9.7(b), and these terms are positive, so Theorem 9.10 gives $a^n = 1/(1/a)^n \to +\infty$.

$a \le -1$: the even terms satisfy $a^{2k} = |a|^{2k} \ge 1$ and the odd terms satisfy $a^{2k+1} \le -1$. A finite limit $L$ would force $L \ge 1$ (from the even terms) and $L \le -1$ (from the odd terms), which is impossible. The limit cannot be $+\infty$ since the odd terms are never positive, nor $-\infty$ since the even terms are never negative. So $\lim a^n$ does not exist.

Tags: StandardLimit laws

Exercise 9.14

Let $p > 0$. Use Exercise 9.12 to show

$$\lim_{n\to\infty} \frac{a^n}{n^p} = \begin{cases} 0 & \text{if } |a| \le 1 \\ +\infty & \text{if } a > 1 \\ \text{does not exist} & \text{if } a < -1. \end{cases}$$

Hint

For $a \ne 0$, $\left|\frac{a^{n+1}/(n+1)^p}{a^n/n^p}\right| = |a|\left(\frac{n}{n+1}\right)^p \to |a|$. For $|a| = 1$ use 9.7(a) directly. For $a > 1$, use Exercise 9.12(b).

Solution

$a = 0$: every term is $0$.

$0 < |a| \le 1$: $\left|\frac{a^n}{n^p}\right| \le \frac1{n^p} \to 0$ (Theorem 9.7(a)), so $\frac{a^n}{n^p} \to 0$ by squeezing.

$|a| > 1$: let $s_n = \frac{a^n}{n^p} \ne 0$. Then $\left|\frac{s_{n+1}}{s_n}\right| = |a|\left(\frac{n}{n+1}\right)^p .$ Choose an integer $k \ge p$. Since $0 < \frac n{n+1} < 1$, we have $\left(\frac n{n+1}\right)^k \le \left(\frac n{n+1}\right)^p \le 1$, and $\left(\frac n{n+1}\right)^k = \left(\frac1{1+1/n}\right)^k \to 1$ by the limit theorems. So the middle term tends to $1$ and the ratio tends to $|a| > 1$. Exercise 9.12(b) gives $|s_n| \to +\infty$. If $a > 1$ the terms are positive, so $s_n \to +\infty$. If $a < -1$, the even terms tend to $+\infty$ and the odd terms to $-\infty$, so the limit does not exist.

Tags: StandardLimit laws

Used in Exercise 9.16

Exercise 9.15

Show $\lim_{n\to\infty} \dfrac{a^n}{n!} = 0$ for all $a \in \mathbb{R}$.

Hint

The ratio of consecutive terms is $\frac{|a|}{n+1} \to 0$. Apply Exercise 9.12(a).

Solution

For $a = 0$ every term is $0$. Otherwise let $s_n = \frac{a^n}{n!} \ne 0$. Then $\left|\frac{s_{n+1}}{s_n}\right| = \frac{|a|}{n+1} \to 0 < 1$, so $s_n \to 0$ by Exercise 9.12(a).

Tags: StandardLimit laws

Used in Exercise 9.16, §16

Exercise 9.16

Use Theorems 9.9 and 9.10 or Exercises 9.9–9.15 to prove the following:

1. $\lim \dfrac{n^4 + 8n}{n^2 + 9} = +\infty$
2. $\lim\left[\dfrac{2^n}{n^2} + (-1)^n\right] = +\infty$
3. $\lim\left[\dfrac{3^n}{n^3} - \dfrac{3^n}{n!}\right] = +\infty$

Solution

1. Write $\dfrac{n^4+8n}{n^2+9} = n^2 \cdot \dfrac{1 + 8/n^3}{1 + 9/n^2}$. Here $n^2 \ge n$, so $n^2 \to +\infty$, and the second factor tends to $1 > 0$. Theorem 9.9 gives $+\infty$.
2. By Exercise 9.14 (with $a = 2$, $p = 2$), $\frac{2^n}{n^2} \to +\infty$. The sequence $(-1)^n$ is bounded, so Exercise 9.11(c) gives $\lim\left[\frac{2^n}{n^2} + (-1)^n\right] = +\infty$.
3. By Exercise 9.14, $\frac{3^n}{n^3} \to +\infty$; by Exercise 9.15, $\frac{3^n}{n!} \to 0$, so $-\frac{3^n}{n!} \to 0 > -\infty$. Exercise 9.11(b) gives the limit $+\infty$.

Tags: StandardLimit laws

Exercise 9.17

Give a formal proof that $\lim n^2 = +\infty$ using only Definition 9.8.

Solution

Let $M > 0$ and $N = \sqrt M$. Then $n > N$ implies $n^2 > M$.

Tags: Warm-upε–N / ε–δ

Exercise 9.18

1. Verify $1 + a + a^2 + \cdots + a^n = \dfrac{1 - a^{n+1}}{1 - a}$ for $a \ne 1$.
2. Find $\lim_{n\to\infty}(1 + a + a^2 + \cdots + a^n)$ for $|a| < 1$.
3. Calculate $\lim_{n\to\infty}\left(1 + \frac13 + \frac19 + \frac1{27} + \cdots + \frac1{3^n}\right)$.
4. What is $\lim_{n\to\infty}(1 + a + a^2 + \cdots + a^n)$ for $a \ge 1$?

Solution

(a) Multiply by $1 - a$; the sum telescopes. (b) $\dfrac1{1-a}$, since $a^{n+1} \to 0$. (c) $\dfrac{1}{1 - 1/3} = \dfrac32$. (d) $+\infty$.

Tags: StandardLimit lawsComputation