1.4 Estimates for Limits

Number Systems, Chapter 4 defined what $a_n \to a$ means in an ordered field (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 4.3.2) and proved the limit laws. It also gave a method for proving a limit from the definition (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Method: proving a limit from the definition)). The method bounds the distance by $C/(n + 1)$ for $n \ge n_0$ , by enlarging numerators and shrinking denominators, and then takes $N \mathrel{:=} \max\{N', n_0\}$ . This section does not teach that method again. It adds two ways to find the bound when the expression is less tame: keep the dominant term of a quotient, and bound a power from below by one term of the binomial theorem. Analysis on the Real Line uses all three.

Throughout, sequences are sequences in $\mathbb{R}$ , a complete ordered field (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.5.2), and $n$ stands for $n \cdot 1$ , as in Number Systems. Where a formula such as $1/n$ needs $n \ge 1$ , the sequence starts at $n = 1$ . Such a sequence is the shift by one of a sequence indexed from $0$ , and a shift changes no limit (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.3.2 (b)).

1.4.1. The method, recalled

The method of Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Method: proving a limit from the definition) has four steps.

(1) Bring the distance $\left| a_n - a \right|$ to one fraction.
(2) Bound it, for every $n \ge n_0$ , by a constant $C > 0$ over $n + 1$ .
(3) Choose a threshold $N'$ for that bound, and let $N \mathrel{:=} \max\{N', n_0\}$ .
(4) Write the proof forwards: $\varepsilon$ first, then $N$ , then an arbitrary $n \ge N$ .

In $\mathbb{R}$ two remarks simplify the work. First, step (3) always succeeds. $\mathbb{R}$ is complete, hence Archimedean (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 6.1.2). So Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.2.3 applies to it.

Second, a bound $C/n$ for $n \ge 1$ serves as well as $C/(n + 1)$ . Since $\mathbb{R}$ is Archimedean, there is an integer $N' > C/\varepsilon$ . Then $1/N' < \varepsilon/C$ , because reciprocals reverse the order of positive numbers (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (b)). So $C/n \le C/N' < \varepsilon$ for every $n \ge N'$ , because a larger denominator gives a smaller quotient (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c)).

Example (A first estimate in

\mathbb{R}

)

Show that $a_n \mathrel{:=} \dfrac{n + 3}{2n^2 - 1} \to 0$ in $\mathbb{R}$ .

Solution. The numbers (1)–(4) are the four steps of the method.

For $n \ge 1$ , $2n^2 - 1 \ge 1 > 0$ and $n + 3 > 0$ , so the distance $\left| a_n - 0 \right|$ is $a_n$ itself.
Let $n \ge 3$ . Then $n + 3 \le n + n = 2n$ . Also $n^2 \ge 1$ , so $2n^2 - 1 = n^2 + (n^2 - 1) \ge n^2 > 0$ . A larger denominator gives a smaller quotient, so $1/(2n^2 - 1) \le 1/n^2$ . Multiplying these two inequalities between non-negative numbers, $0 < a_n \le 2n \cdot \frac{1}{n^2} = \frac{2}{n} \qquad (n \ge 3) .$ So $n_0 = 3$ and $C = 2$ .
Let $\varepsilon > 0$ . Choose an integer $N' > 2/\varepsilon$ and let $N \mathrel{:=} \max\{3, N'\}$ . It depends on $\varepsilon$ only.
Let $n \ge N$ . Then $n \ge 3$ , so the bound of (2) applies, and $2/n \le 2/N$ since $n \ge N$ : $\left| a_n - 0 \right| = a_n \le \frac{2}{n} \le \frac{2}{N} < \varepsilon ,$ where the last step uses $N \ge N' > 2/\varepsilon$ . Hence $a_n \to 0$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c) (larger denominator, smaller quotient), (d) (multiplying inequalities).

Both bounds in step (2) needed a range. The bound $n + 3 \le 2n$ fails at $n = 2$ . The bound $2n^2 - 1 \ge n^2$ holds from $n = 1$ on. Here the ranges were found at sight. When the terms do not simplify so readily, two techniques find them: keeping the dominant term, and bounding a power from below.

1.4.2. Keeping the dominant term

Take $\dfrac{2n^2 + 3n}{n^2 - 5}$ . For large $n$ the terms $2n^2$ and $n^2$ swamp the others, and the quotient should be close to $2$ . To bound a quotient from above we need its numerator from above and its denominator from below. Enlarging a numerator is easy. Shrinking a denominator that contains a subtraction is where care is needed, because $n^2 - 5$ is negative for small $n$ .

Example (Method: keep the dominant term)

How it goes. In a quotient of sums, keep the term that grows fastest and absorb the others into it. From some $n_0$ on, a lower-order term is at most a fixed fraction of the dominant term. For example, $5 \le \frac12 n^2$ once $n \ge 4$ , so $n^2 - 5 \ge \frac12 n^2$ there. Then the quotient is bounded by a quotient of single powers. This is the bounding step of the method of Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Method: proving a limit from the definition), with $n_0$ the largest of the ranges used.

When to use it. Rational expressions in $n$ , possibly with $\sqrt{n}$ .

Common mistakes. Subtracting in the denominator without stating a range. For instance, $n^2 - 5 \ge \frac12 n^2$ is false for $n \le 3$ .

Example (A limit that is not zero)

Show that $a_n \mathrel{:=} \dfrac{2n^2 + 3n}{n^2 - 5} \to 2$ .

Solution.

No natural number has square $5$ , so every $a_n$ is defined. For $n \ge 3$ , $n^2 \ge 9 > 5$ , and $a_n - 2 = \frac{2n^2 + 3n - 2(n^2 - 5)}{n^2 - 5} = \frac{3n + 10}{n^2 - 5} > 0 ,$ so the distance is $\left| a_n - 2 \right| = \dfrac{3n + 10}{n^2 - 5}$ .
Keep the dominant terms, $3n$ on top and $n^2$ below. Let $n \ge 4$ . Then $3n \ge 12 \ge 10$ , so $3n + 10 \le 6n$ . And $\frac12 n^2 \ge \frac12 \cdot 16 = 8 \ge 5$ , so $n^2 - 5 \ge \frac12 n^2 > 0$ . Inverting the second, which reverses it, and multiplying by the first, $\left| a_n - 2 \right| \le 6n \cdot \frac{2}{n^2} = \frac{12}{n} \qquad (n \ge 4) .$
Let $\varepsilon > 0$ . Choose an integer $N' > 12/\varepsilon$ and let $N \mathrel{:=} \max\{4, N'\}$ . It depends on $\varepsilon$ only.
For $n \ge N$ : $\left| a_n - 2 \right| \le 12/n \le 12/N < \varepsilon$ . Hence $a_n \to 2$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c) (larger denominator, smaller quotient), (d) (multiplying inequalities).

The candidate limit $2$ is the ratio of the leading coefficients. Reading off the dominant terms is how the limit is guessed, and the method is how the guess is proved. The range $n \ge 4$ came from two requirements, $3n \ge 10$ and $\frac12 n^2 \ge 5$ , and the larger of their thresholds serves both.

Quick check. For which $n \in \mathbb{N}$ is $n^2 - 7n \ge \frac12 n^2$ ?

Solution. The inequality says $\frac12 n^2 \ge 7n$ , that is, $n \bigl( \frac12 n - 7 \bigr) \ge 0$ . For $n = 0$ it holds with equality. For $n \ge 1$ the factor $n$ is positive, so it holds exactly when $\frac12 n \ge 7$ , that is, $n \ge 14$ . So the inequality holds for $n = 0$ and for every $n \ge 14$ , and fails for $1 \le n \le 13$ . A proof that uses $n^2 - 7n \ge \frac12 n^2$ takes $n_0 = 14$ .

1.4.3. Powers

Every bound so far came from dropping or absorbing terms of a sum. A power $r^n$ has only one term, and there is nothing to drop. Bernoulli’s inequality and the binomial theorem turn it into a sum, and a sum of non-negative terms is at least any one of its terms.

The first case is the geometric sequence. For $\left| r \right| \ge 1$ the terms $r^n$ do not shrink, and for $\left| r \right| < 1$ they should tend to $0$ . Take $r = \frac12$ first. Then $(\frac12)^n = 1/2^n$ , and Bernoulli’s inequality gives $2^n = (1 + 1)^n \ge 1 + n > n$ . So $(\frac12)^n < 1/n$ for every $n \ge 1$ , a bound of the form $C/n$ with $C = 1$ . The proposition below does the same for every $r$ with $0 < \left| r \right| < 1$ , by writing $1/\left| r \right|$ as $1 + h$ .

Proposition 1.4.1 (Geometric Sequences Tend to Zero)

Let $r \in \mathbb{R}$ with $\left| r \right| < 1$ .

(a) If $r \ne 0$ and $h \mathrel{:=} \dfrac{1}{\left| r \right|} - 1$ , then $h > 0$ and $\left| r \right|^n < \dfrac{1}{nh}$ for every integer $n \ge 1$ .
(b) $r^n \to 0$ .

Idea (Scratch work). The goal, from Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 4.3.2, is $\left| r^n - 0 \right| = \left| r \right|^n < \varepsilon$ for every $n$ beyond some $N$ . Solving $\left| r \right|^n < \varepsilon$ for $n$ would need a logarithm, which we do not have yet. So we follow the bounding step of the method and look for a simpler bound of the form $C/n$ .

The trouble is that $\left| r \right|^n$ is a small number. The tools we have, Bernoulli’s inequality and the binomial theorem, bound powers of numbers above $1$ from below. So we turn small into large: we write $\left| r \right|$ as the reciprocal of a number above $1$ , $\left| r \right| = \frac{1}{1 + h} , \qquad h \mathrel{:=} \frac{1}{\left| r \right|} - 1 > 0 ,$ so that $\left| r \right|^n = 1/(1 + h)^n$ . Now $\left| r \right|^n$ is small as soon as $(1 + h)^n$ is large, and a lower bound $(1 + h)^n \ge Cn$ is enough. Bernoulli’s inequality supplies one at once: $(1 + h)^n \ge 1 + nh > nh$ .

We know	We want
$0 < \left\| r \right\| < 1$	$\left\| r \right\|^n$ below every $\varepsilon > 0$ for large $n$
$h \mathrel{:=} \frac{1}{\left\| r \right\|} - 1 > 0$ , so $\left\| r \right\|^n = \frac{1}{(1 + h)^n}$	enough: $(1 + h)^n$ bounded below by $Cn$
Sets and Functions, Theorem 3.3.7: $(1 + h)^n \ge 1 + nh > nh$	$\left\| r \right\|^n < \frac{1}{nh}$ : the bounding step of the method with $C = \frac{1}{h}$ ; the gap is closed

The proof follows these rows from top to bottom. It then finishes with the choice of the threshold and the forward proof, using the bound $1/(nh)$ .

Proof.

(a) Let $r \ne 0$ . Then $0 < \left| r \right| < 1$ , since the absolute value of a non-zero number is positive. Reciprocals reverse the order of positive numbers, so $1/\left| r \right| > 1/1 = 1$ , that is, $h > 0$ . By the choice of $h$ , $1 + h = 1/\left| r \right|$ , so $\left| r \right| = 1/(1 + h)$ and $\left| r \right|^n = 1/(1 + h)^n$ . Since $h \ge -1$ , Bernoulli’s inequality (Sets and Functions, Theorem 3.3.7) gives $(1 + h)^n \ge 1 + nh > nh .$ For $n \ge 1$ , $nh > 0$ as a product of positive numbers. So $0 < nh < (1 + h)^n$ , and taking reciprocals reverses the inequality: $\left| r \right|^n = 1/(1 + h)^n < 1/(nh)$ .
(b) If $r = 0$ , then $r^n = r^{n-1} \cdot 0 = 0$ for every $n \ge 1$ , so $\left| r^n - 0 \right| = 0 < \varepsilon$ for every $\varepsilon > 0$ and every $n \ge 1$ . So $N \mathrel{:=} 1$ works for every $\varepsilon > 0$ . Let $r \ne 0$ and $\varepsilon > 0$ , and let $h$ be as in (a). Then $h\varepsilon > 0$ . Since $\mathbb{R}$ is Archimedean, Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.2.3 gives $N \in \mathbb{Z}_{>0}$ with $1/n < h\varepsilon$ for every $n \ge N$ . This $N$ depends on $\varepsilon$ and on $r$ , through $h$ . Let $n \ge N$ . Then $n \ge 1$ , and $\left| r^n \right| = \left| r \right|^n$ by induction on $n$ , since $\left| xy \right| = \left| x \right| \left| y \right|$ . By (a), $\left| r^n - 0 \right| = \left| r \right|^n < \frac{1}{nh} < \frac{h\varepsilon}{h} = \varepsilon .$ This proves $r^n \to 0$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (a) ( $\left| r \right| > 0$ for $r \ne 0$ ), (b) ( $\left| xy \right| = \left| x \right| \left| y \right|$ ); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (b) (reciprocals reverse $<$ ). Sets and Functions, Proposition 3.3.4 ( $(1/x)^n = 1/x^n$ ); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 4.2.1 (b) ( $x^n = x^{n-1} x$ ); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (products of positives).

Part (a) is the useful half. It gives an explicit bound, valid for every $n \ge 1$ , from which a threshold can be read off for any tolerance.

Quick check. From Proposition 1.4.1 (a), find an $N$ with $(0.9)^n < 0.01$ for every $n \ge N$ .

Solution. Here $r = 0.9 = \frac{9}{10}$ , so $h = \frac{10}{9} - 1 = \frac{1}{9}$ and $(0.9)^n < \dfrac{1}{n/9} = \dfrac{9}{n}$ for $n \ge 1$ . Now $\frac{9}{n} \le \frac{1}{100}$ exactly when $n \ge 900$ . So $N = 900$ works: for $n \ge 900$ , $(0.9)^n < 9/n \le 0.01$ . The bound is wasteful, since already $(0.9)^{44} < 0.01$ . That costs nothing: the definition asks for some threshold, not the least.

The same steps handle a power in a denominator under a numerator that is a power of $n$ : write the base as $1 + h$ , keep one term of the expansion, and invert.

Example (Method: bound a power from below)

How it goes. To make a quotient small when its denominator contains $r^n$ with $r > 1$ :

(1) Write $r = 1 + h$ with $h > 0$ .
(2) Keep one term of the expansion: $(1 + h)^n \ge nh$ by Bernoulli’s inequality, or $(1 + h)^n \ge \binom{n}{k} h^k$ by Corollary 1.2.4 (b). Take $k$ one more than the power of $n$ on top.
(3) Bound $\binom{n}{k}$ below by $c\,n^k$ for $n \ge n_0$ .
(4) Invert, and finish with the method of Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Method: proving a limit from the definition).

Common mistakes. One mistake is to choose $k$ too small: then the bound does not tend to $0$ . Another is to use the binomial bound with $h < 0$ .

In Example (A lower bound for powers of two) the base was $2$ , and $k = 2$ gave $n/2^n \le 2/(n - 1)$ for $n \ge 2$ . The same steps work for every base $r > 1$ . The numerator is $n$ , to the first power, so $k = 2$ .

Proposition 1.4.2 (Powers Beat n)

Let $r > 1$ and $h \mathrel{:=} r - 1$ . Then $\dfrac{n}{r^n} \le \dfrac{4}{nh^2}$ for every integer $n \ge 2$ , and $\dfrac{n}{r^n} \to 0$ .

Idea. Follow Example (Method: bound a power from below) step by step. Write $r = 1 + h$ . Keep the term $k = 2$ , which is of order $n^2$ . Bound $n(n - 1)$ below by $\frac12 n^2$ . Invert, multiply by $n$ , and finish the limit with $C = 4/h^2$ .

Proof. Let $n \ge 2$ . Since $r > 1$ , $h > 0$ . By Corollary 1.2.4 (b) with $x = h$ and $k = 2 \le n$ , and since $\binom{n}{2} \cdot 2 = n(n - 1)$ , $r^n = (1 + h)^n \ge \binom{n}{2} h^2 = \frac{n(n - 1)}{2} h^2 .$ Since $n \ge 2$ , $\frac{n}{2} \ge 1$ , so $n - 1 \ge \frac{n}{2} \ge 0$ . Multiplying the inequalities $0 \le \frac{n}{2} \le n - 1$ and $0 \le n \le n$ , between non-negative numbers, gives $n(n - 1) \ge \frac{n^2}{2}$ . Hence $r^n \ge \frac{n^2 h^2}{4} > 0$ . Inverting, which reverses the inequality, and multiplying by $n > 0$ , $\frac{n}{r^n} \le n \cdot \frac{4}{n^2 h^2} = \frac{4}{nh^2} .$ For the limit, $n/r^n > 0$ , so the distance to $0$ is $n/r^n \le C/n$ for $n \ge 2$ , with $C \mathrel{:=} 4/h^2$ . Let $\varepsilon > 0$ , choose an integer $N' > C/\varepsilon$ , and let $N \mathrel{:=} \max\{2, N'\}$ . It depends on $\varepsilon$ and $r$ . For $n \ge N$ , $\left| n/r^n - 0 \right| \le C/n \le C/N < \varepsilon$ . So $n/r^n \to 0$ .

Uses: Proposition 1.2.2 (c) ( $\binom{n}{2} \cdot 2 = n(n - 1)$ ); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c) (larger denominator, smaller quotient), (d) (multiplying inequalities).

With $k = 1$ , that is, Bernoulli’s inequality, the same steps give only $n/r^n < n/(nh) = 1/h$ . That bound does not tend to $0$ . The choice of $k$ is the whole difference.

A comparison with a power can also run the other way: a sequence that is hard to handle directly is bounded by a geometric one, and Proposition 1.4.1 finishes.

Example (Factorials beat powers of two)

Show that $\dfrac{1}{n!} \le \dfrac{1}{2^{n-1}} < \dfrac{2}{n}$ for every integer $n \ge 1$ , and hence that $\dfrac{1}{n!} \to 0$ .

Solution. The first inequality. We show $n! \ge 2^{n-1}$ for $n \ge 1$ , by induction on $n$ . For $n = 1$ : $1! = 1 = 2^0$ . Suppose $n! \ge 2^{n-1}$ . Also $n + 1 \ge 2$ , and both inequalities are between non-negative numbers, so we may multiply them: $(n + 1)! = n! \, (n + 1) \ge 2^{n-1} \cdot 2 = 2^n .$ Here $2^{n-1} > 0$ , as a power of a positive number, and $n! \ge 2^{n-1} > 0$ . So inverting reverses the inequality, and $1/n! \le 1/2^{n-1}$ .

The second inequality. By Proposition 1.4.1 (a) with $r = \frac12$ , $h = 2 - 1 = 1$ , we have $(\frac12)^n < \frac1n$ for $n \ge 1$ . Since $1/2^{n-1} = (\frac12)^{n-1} = 2 \cdot (\frac12)^n$ , $\frac{1}{2^{n-1}} = 2 \Bigl( \frac12 \Bigr)^n < \frac{2}{n} .$

The limit. So $0 < 1/n! < 2/n$ for $n \ge 1$ : a bound $C/n$ with $C = 2$ . Given $\varepsilon > 0$ , take an integer $N > 2/\varepsilon$ . It depends on $\varepsilon$ only. Then $N \ge 1$ , and for $n \ge N$ , $\left| 1/n! - 0 \right| < 2/n \le 2/N < \varepsilon$ . Hence $1/n! \to 0$ .

Uses: Sets and Functions, Definition 3.3.1 ( $(n + 1)! = n! \, (n + 1)$ ); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (d) (multiplying inequalities), (c) (larger denominator, smaller quotient); Sets and Functions, Proposition 3.3.4 ( $1/2^{n-1} = 2 (\frac12)^n$ ), (e) ( $2^{n-1} > 0$ ).

Warning. $\frac{1}{n^2} \le \varepsilon$ is not $\frac{1}{n^2} < \varepsilon$ . The definition of a limit asks for the strict inequality. A threshold chosen to make the bound at most $\varepsilon$ can miss it. For $\varepsilon = 1$ , the choice $N \mathrel{:=} 1/\varepsilon = 1$ gives, at $n = N = 1$ , the distance $\frac{1}{1^2} = 1$ , which is not $< 1$ . Ask for $N > C/\varepsilon$ , strictly.

Second readingWhen the one-term bound fails

Change one hypothesis of Corollary 1.2.4 (b): keep $k = 2$ and allow $h < 0$ . At $h = -1$ , $n = 2$ the claim would read $(1 - 1)^2 \ge \binom{2}{2}(-1)^2$ , that is, $0 \ge 1$ , which is false. What breaks is the step that keeps one term of a sum. That step needs every term non-negative, and with $h < 0$ the terms $\binom{n}{j} h^j$ alternate in sign. Bernoulli’s inequality survives for $-1 \le h < 0$ , because its proof is an induction, not a dropped sum. So for $0 < \left| r \right| < 1$ the method first passes to $1/\left| r \right| = 1 + h$ with $h > 0$ , as the proof of Proposition 1.4.1 did. It never expands $(1 + (r - 1))^n$ directly.

1.4.4. Exercises

A. Check your understanding

Exercise (A1)

(a) Write the negation of “ $a_n \to 0$ ” with all its quantifiers.
(b) In Example (A first estimate in $\mathbb{R}$ ), on what does $N$ depend? Could it depend on $n$ ?
(c) Name the method used: $\dfrac{n}{n^3 + 7} \le \dfrac{1}{n^2}$ for $n \ge 1$ .
(d) True or false: if $\left| a_n \right| \le \dfrac{5}{n}$ for every $n \ge 10$ , then $a_n \to 0$ .

Solution.

(a) $\exists \varepsilon > 0 \ \ \forall N \in \mathbb{N} \ \ \exists n \in \mathbb{N} \ \ \bigl( n \ge N \wedge \left| a_n \right| \ge \varepsilon \bigr)$ , as in the negation after Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 4.3.2 with $a = 0$ .
(b) On $\varepsilon$ only. It cannot depend on $n$ : $N$ is chosen before $n$ is named, and the claim is about every $n \ge N$ .
(c) Bounding the distance by shrinking the denominator, the bounding step of Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (Method: proving a limit from the definition). Here we keep the dominant term $n^3$ and drop the positive $7$ , as in Example (Method: keep the dominant term). So $n^3 + 7 \ge n^3 > 0$ and $n/(n^3 + 7) \le n/n^3 = 1/n^2$ .
(d) True. Given $\varepsilon > 0$ , take an integer $N' > 5/\varepsilon$ and $N \mathrel{:=} \max\{10, N'\}$ . For $n \ge N$ , $\left| a_n - 0 \right| \le 5/n \le 5/N < \varepsilon$ .

B. Practice

Exercise (B1: Say which method, then complete it)

Show that $a_n \mathrel{:=} \dfrac{n + 5}{2^n} \to 0$ . Start: write $2 = 1 + 1$ . By Corollary 1.2.4 (b) with $k = 2$ , $2^n \ge \dfrac{n(n - 1)}{2} \ge \dfrac{n^2}{4}$ for $n \ge 2$ . Say which method this is, then complete it.

Solution. The method: bound the power from below, as in Example (Method: bound a power from below). The numerator is handled by keeping its dominant term, as in Example (Method: keep the dominant term).

Let $n \ge 5$ . Then $n + 5 \le 2n$ . By the start, $2^n \ge n^2/4 > 0$ , so $1/2^n \le 4/n^2$ . Multiplying, $0 < a_n \le 2n \cdot \frac{4}{n^2} = \frac{8}{n} .$ Let $\varepsilon > 0$ , take an integer $N' > 8/\varepsilon$ and $N \mathrel{:=} \max\{5, N'\}$ . For $n \ge N$ , $\left| a_n - 0 \right| \le 8/n \le 8/N < \varepsilon$ . Hence $a_n \to 0$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c) (larger denominator, smaller quotient), (d) (multiplying inequalities).

Exercise (B2: Put the proof in order)

The claim is $a_n \mathrel{:=} \dfrac{2n + 1}{n + 1} \to 2$ . Put the lines below in an order that proves it, and say which line does not belong.

(i) Let $\varepsilon > 0$ .
(ii) For $n \ge N$ , $\left| a_n - 2 \right| = \dfrac{1}{n + 1} < \dfrac{1}{n} \le \dfrac{1}{N} < \varepsilon$ .
(iii) Choose $N \in \mathbb{N}$ with $N > 1/\varepsilon$ and $N \ge 1$ .
(iv) Choose $N \mathrel{:=} n + 1$ .
(v) Hence $a_n \to 2$ .
(vi) $\left| a_n - 2 \right| = \Bigl\lvert \dfrac{2n + 1 - 2n - 2}{n + 1} \Bigr\rvert = \dfrac{1}{n + 1}$ for every $n \in \mathbb{N}$ .

Solution. An admissible order is (i), (iii), (vi), (ii), (v). Line (vi) uses nothing chosen, so it may also come first: (vi), (i), (iii), (ii), (v). Line (ii) must come after (iii), since it uses $N$ . It must also come after (vi), since it uses the formula for the distance. In (ii), $1/n \le 1/N$ holds because a larger denominator gives a smaller quotient, and $1/N < \varepsilon$ follows from $N > 1/\varepsilon$ .

The line that does not belong is (iv): it makes $N$ depend on $n$ , but $N$ must be fixed before $n$ is named, and $n$ ranges over all numbers $\ge N$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c) (larger denominator, smaller quotient); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (b) (reciprocals reverse $<$ ).

Exercise (B3: Prove or disprove)

Prove each true statement and refute each false one. In parts (a), (b), (c) and (e), $a_n$ denotes the $n$ -th term of the sequence in question.

(a) $\dfrac{n^2 + 1}{n^3} \to 0$ .
(b) $\dfrac{n}{n + 1} \to 0$ .
(c) $\dfrac{(-1)^n}{n} \to 0$ .
(d) For every $\varepsilon > 0$ , every integer $n \ge 1/\varepsilon$ has $\dfrac{1}{n^2} < \varepsilon$ .
(e) $\dfrac{2^n}{n!} \to 0$ .

Solution.

(a) True. For $n \ge 1$ , $n^2 + 1 \le 2n^2$ , so $0 < a_n \le 2n^2/n^3 = 2/n$ . Given $\varepsilon > 0$ , take an integer $N > 2/\varepsilon$ . Then $N \ge 1$ , and for $n \ge N$ , $\left| a_n \right| \le 2/n \le 2/N < \varepsilon$ .
(b) False. For $n \ge 1$ , $2n \ge n + 1$ , so $\frac{n}{n + 1} \ge \frac12$ . Take $\varepsilon \mathrel{:=} \frac12$ . For every $N \in \mathbb{N}$ , the index $n \mathrel{:=} \max\{N, 1\} \ge N$ has $\left| a_n - 0 \right| \ge \frac12 = \varepsilon$ . This is the negation of A1 (a).
(c) True. $\left| a_n - 0 \right| = \left| (-1)^n \right|/n = 1/n$ , since $\left| xy \right| = \left| x \right| \left| y \right|$ and $\left| (-1)^n \right| = 1$ . Given $\varepsilon$ , an integer $N > 1/\varepsilon$ works, as in (a).
(d) False. Take $\varepsilon = 1$ and $n = 1 \ge 1/\varepsilon$ . Then $1/n^2 = 1$ , which is not $< 1$ . The non-strict $1/n^2 \le \varepsilon$ does not give the strict $1/n^2 < \varepsilon$ .
(e) True. For $n \ge 3$ , $n! = 1 \cdot 2 \cdot \prod_{k=3}^{n} k$ . Also $2^n/n! = \prod_{k=1}^{n} (2/k)$ , by induction on $n$ from $2^{n+1}/(n+1)! = (2^n/n!) \cdot (2/(n+1))$ . So $\frac{2^n}{n!} = \frac{2}{1} \cdot \frac{2}{2} \cdot \prod_{k=3}^{n} \frac{2}{k} \le 2 \cdot 1 \cdot \frac{2}{n} = \frac{4}{n} .$ Every factor $\frac{2}{k}$ with $3 \le k < n$ lies in $(0, 1]$ , so their product lies in $(0, 1]$ . The last factor, $k = n$ , is kept. Given $\varepsilon > 0$ , take an integer $N' > 4/\varepsilon$ and $N \mathrel{:=} \max\{3, N'\}$ . For $n \ge N$ , $\left| a_n \right| \le 4/n \le 4/N < \varepsilon$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (b) ( $\left| xy \right| = \left| x \right| \left| y \right|$ ); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Example (First Limits) (d) ( $\left| (-1)^n \right| = 1$ ); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (d) (multiplying inequalities); Sets and Functions, Definition 3.3.1 (the recursion of products and factorials, for $2^n/n! = \prod_{k=1}^n (2/k)$ and the splitting of $n!$ ).

C. Going deeper

Exercise (C1: Chapter check: a rational limit)

Prove from the definition that $a_n \mathrel{:=} \dfrac{3n^2 - n}{n^2 + 4} \to 3$ .

Solution. Let $\varepsilon > 0$ . The goal is an $N$ with $\left| a_n - 3 \right| < \varepsilon$ for every $n \ge N$ .

The distance. For every $n$ , $a_n - 3 = \frac{3n^2 - n - 3(n^2 + 4)}{n^2 + 4} = \frac{-(n + 12)}{n^2 + 4} ,$ so $\left| a_n - 3 \right| = \dfrac{n + 12}{n^2 + 4}$ .

The bound. Let $n \ge 1$ . Then $n + 12 \le n + 12n = 13n$ , and $n^2 + 4 \ge n^2 > 0$ , so $1/(n^2 + 4) \le 1/n^2$ . Multiplying, $\left| a_n - 3 \right| \le 13n/n^2 = 13/n$ .

The threshold. Choose an integer $N > 13/\varepsilon$ . Then $N \ge 1$ , and $N$ depends on $\varepsilon$ only. For $n \ge N$ , $n \ge 1$ , so $\left| a_n - 3 \right| \le \frac{13}{n} \le \frac{13}{N} < \varepsilon ,$ where the middle step holds because $n \ge N$ and the last because $N > 13/\varepsilon$ . Hence $a_n \to 3$ .

What a complete answer needs. (10 points)

2: fixes $\varepsilon > 0$ arbitrary and states the goal.
3: computes $\left| a_n - 3 \right|$ exactly.
3: bounds it by $C/n$ (or $C/(n + 1)$ ) on a stated range $n \ge n_0$ , with each inequality justified.
2: chooses $N$ before $n$ , depending only on $\varepsilon$ , and ends the chain with a strict $< \varepsilon$ .

Slips that lose the step. Letting $N$ depend on $n$ . Bounding in the wrong direction, that is, shrinking a numerator or enlarging a denominator.

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c) (larger denominator, smaller quotient), (d) (multiplying inequalities).

Exercise (C2: Squares over powers)

Let $r > 1$ and $h \mathrel{:=} r - 1$ . Show that $\dfrac{n^2}{r^n} \to 0$ . Start: by Corollary 1.2.4 (b) with $k = 3$ , $r^n \ge \dfrac{n(n - 1)(n - 2)}{6} h^3$ for $n \ge 3$ . Say which method this is, then complete it.

Solution (Hint 1). Bound $n - 1$ and $n - 2$ below by multiples of $n$ , for $n \ge 4$ .

Solution (Hint 2). For $n \ge 4$ , $n - 2 \ge \frac{n}{2}$ , since this rearranges to $\frac{n}{2} \ge 2$ . Find a similar bound for $n - 1$ , then multiply.

Solution. The method: bound the power from below, as in Example (Method: bound a power from below). The start wrote $r = 1 + h$ and kept one term, with $k = 3$ , one more than the power of $n$ on top. What remains is to bound $n(n - 1)(n - 2)$ below and finish the limit.

The lower bound. Let $n \ge 4$ . Then $n - 1 \ge \frac{3n}{4}$ , since this rearranges to $\frac{n}{4} \ge 1$ . Also $n - 2 \ge \frac{n}{2}$ , since this rearranges to $\frac{n}{2} \ge 2$ . Both sides of both inequalities are non-negative. So we may multiply them, with $0 \le n \le n$ as the third factor, and get $n(n - 1)(n - 2) \ge n \cdot \frac{3n}{4} \cdot \frac{n}{2} = \frac{3n^3}{8}$ . Hence $r^n \ge \frac{3n^3}{8 \cdot 6} h^3 = \frac{n^3 h^3}{16} > 0 \qquad (n \ge 4) .$

The limit. Inverting, which reverses the inequality, and multiplying by $n^2 > 0$ , we get $\dfrac{n^2}{r^n} \le \dfrac{16}{nh^3}$ for $n \ge 4$ . Let $\varepsilon > 0$ , take an integer $N' > \dfrac{16}{h^3 \varepsilon}$ and $N \mathrel{:=} \max\{4, N'\}$ . It depends on $\varepsilon$ and $r$ . For $n \ge N$ , $\Bigl\lvert \frac{n^2}{r^n} - 0 \Bigr\rvert \le \frac{16}{nh^3} \le \frac{16}{Nh^3} < \varepsilon .$ Hence $n^2/r^n \to 0$ .

What a complete answer needs. (6 points)

1: names the method (bounding a power from below).
2: linear lower bounds for $n - 1$ and $n - 2$ on a stated, justified range $n \ge n_0$ .
1: their product, with non-negativity checked, and the inversion, with positivity stated.
2: the choice of $N$ , and the strict chain ending in $< \varepsilon$ .

Slips that lose the step. Using $n - 1 \ge \frac{3n}{4}$ or $n - 2 \ge \frac{n}{2}$ without the range $n \ge 4$ . Multiplying inequalities without checking that the factors are non-negative.

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (d) (multiplying inequalities, twice), (c) (larger denominator, smaller quotient).