[Understanding Analysis by Stephen Abbott] - [Chapter 2 Sequences and Series] - [2.3 The Algebraic and Order Limit Theorems]

The real purpose of creating a rigorous definition for convergence of a sequence is not to have a tool to verify computational statements such as $\lim 2n/(n+2) = 2$. The point of having such a logically tight description of convergence is so that we can confidently prove statements about convergent sequences in general. We are ultimately trying to resolve arguments about what is and is not true regarding the behavior of limits with respect to the mathematical manipulations we intended to inflict on them.

Definition 2.3.1 A sequence $(x_n)$ is bounded if there exists a number $M > 0$ such that $|x_n| \leqslant M$ for all $n \in N$

Geometrically, this means that we can find an interval $[-M, M]$ that contains every term in the sequence $(x_n)$

Theorem 2.3.2 Every convergent sequence is bounded.Every convergent sequence is bounded.

Proof. Assume $(x_n)$ converges to a limit $l$. This means that given a particular value of $\epsilon$, say $\epsilon = 1$, we know there must exist an $N in N$ such that if $n \geqslant N$, then $x_n$ is in the interval $(l - 1, l + 1)$. Not knowing whether $l$ is positive or negative, we can certainly conclude that $$ |x_n| < |l| + 1 $$ for all $n \geqslant N$

We still need to worry about the terms in the sequence that come before the $Nth$ term. Because there are only a finite number of these, we let $$ M = max\{|x_1|, |x_2|, |x_3|,...,|x_{N - 1}|, |l| + 1\} $$ It follows that $|x_n| \leqslant M$ for all $n \in N$.

Theorem 2.3.3 Algebraic Limit Theorem Let $\lim a_n = a$, and $\lim b_n = b$. Then,

$\lim(ca_n) = ca$, for all $c \in R$
$\lim(a_n + b_n) = a + b$
$\lim(a_n b_n) = ab$
$\lim(a_n/b_n) = a/b$, provided $b \neq 0$

Proof.

Consider the case where $c \neq 0$. We want to show that the sequence $(ca_n)$ converges to$ca$. Let $\epsilon$ be some arbitrary positive number. Our goal is to find some point in the sequence $ca_n$ after which we have $$ |ca_n - ca| < \epsilon $$ Now, $$ |ca_n - ca| = |c||a_n - a| $$ We are given that $(a_n) \to a$, so we know we can make $|a_n - a|$ as small as we like. We can choose an $N$ such that $$ |a_n - a| < \frac{\epsilon}{|c|} $$ whenever $n \geqslant N$. To see that this $N$ indeed works, observe that, for all $n \geqslant N$, $$ |ca_n - ca| = |c||a_n - a| < |c|\frac{\epsilon}{|c|} = \epsilon $$ The case $c = 0$ reduces to showing that the constant sequence $(0,0,0,...)$ converges to $0$.
To prove this statement, we need to argue that the quantity $$ |(a_n + b_n) - (a + b)| $$ can be made less than an arbitrary $\epsilon$ using the assumption that $|a_n - a|$ and $|b_n - b|$ can be made as small as we like for large $n$. The first step is to use the triangle inequality to say $$ |(a_n + b_n) - (a + b)| = |(a_n - a) + (b_n - b)| \leqslant |a_n - a| + |b_n - b| $$ Again, we let $\epsilon > 0$ be arbitrary. The technique this time is to divide the $\epsilon$ between the two expressions on the right-hand side in the preceding inequality. Using the hypothesis that $(a_n) \to a$, we know there exits an $N_1$ such that $$ |a_n - a| < \frac{\epsilon}{2} \text{ whenever } n \geqslant N_1 $$ Likewise, the assumption that $(b_n) \to b$ means that we can choose an $N_2$ so that $$ |b_n - b| < \frac{\epsilon}{2} \text{ whenever } n \geqslant N_2 $$ By choosing $N = max\{N_1, N_2\}$, we get $$ \begin{align} |(a_n + b_n) - (a + b)| &\leqslant |a_n - a| + |b_n - b| \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align} $$ for all $n \geqslant N$, as desired.
To show that $(a_nb_n) \to ab$, we begin by observing that $$ \begin{align} |a_nb_n - ab| &= |a_nb_n - ab_n + ab_n - ab| \\ &= |a_nb_n - ab_n| + |ab_n - ab| \\ &= |b_n||a_n - a| + |a||b_n - b| \end{align} $$
In the initial step, we subtracted and then added $ab_n$, wich created an opportunity to use the triangle inequality. Essentially, we have broken up the distance from $a_nb_n$ to $ab$ with a midway point and are using the sum of the two distances to overestimate the original distance.

Letting $\epsilon > 0$ be arbitrary, we again proceed with the strategy of making each piece in the preceding inequality less than $\epsilon/2$. For the piece on the right-hand side $(|a||b_n - b|)$, if $a \neq 0$ we can choose $N_1$ so that $$ n \geqslant N_1 \text{ implies } |b_n - b| < \frac{1}{|a|}\frac{\epsilon}{2} $$ Getting the term on left-hand side $|b_n||a_n - a|$ to be less than $\epsilon/2$ is complicated by the fact that we have a variable quantity $|b_n|$ to contend with as opposed to the constant $|a|$ we encountered in the right-hand term. The idea is to replace $|b_n|$ with a worst-case estimate. Using the fact that convergent sequences are bounded, we know there exists a bound $M > 0$ satisfying $|b_n| \leqslant M$ for all $n \in M$. Now we can choose $N_2$ so that $$ |a_n - a| < \frac{1}{M}\frac{\epsilon}{2} \text{ whenever } n \geqslant N_2 $$ To finish the argument, pick $N = max{N_1, N_2}$, and observe that if $n \geqslant N$, then $$ \begin{align} |a_nb_n - ab| &\leqslant |a_nb_n - ab_n| + |ab_n - ab| \\ &= |b_n||a_n - a| + |a||b_n - b| \\ &\leqslant M|a_n - a| + |a||b_n - b| \\ &< M(\frac{\epsilon}{M2}) + |a|(\frac{\epsilon}{|a|2}) = \epsilon \end{align} $$
This final statement will follow from previous proof if we can prove that $$ (b_n) \to b \text{ implies } (\frac{1}{b_n}) \to \frac{1}{b} $$ whenever $b \neq 0$. We have $$ |\frac{1}{b_n} - \frac{1}{b}| = \frac{|b - b_n|}{|b||b_n|} $$ Because $(b_n) \to b$, we can make the preceding numerator as small as we like by choosing $n$ large. The problem comes in that we need a worst-case estimate on the size of $1/(|b||b_n|)$. Because the $b_n$ terms are in the denominator, we are no longer interested in an upper bound on |b_n| but rather in an inequality of the form $|b_n| \geqslant \delta > 0$. This will then lead to a bound on the size of $1/(|b||b_n|)$
The trick is to look far enough out into the sequence $b_n$ so that the terms are closer to b than they are to 0. Consider the particular value $\epsilon_0 = |b|/2$. Because $(b_n) \to b$, there exists an $N_1$ such that $|b_n - b| < |b|/2$ for all $n \geqslant N_1$. This implies $|b_n| > |b|/2$

Next, choose $N_2$ so that $n \geqslant N_2$ implies $$ |b_n - b| < \frac{\epsilon|b|^2}{2} $$

Finally, if we let $N = max\{N_1, N_2\}$, then $n \geqslant N$ implies $$ |\frac{1}{b_n} - \frac{1}{b}| = |b - b_n|\frac{1}{|b||b_n|} < \frac{\epsilon|b|^2}{2}\frac{1}{|b|\frac{|b|}{2}} = \epsilon $$

Limits and Order

Theorem 2.3.4 (Order Limit Theorem) Assume $\lim a_n = a$ and $\lim b_n = b$

If $a_n > 0$ for all $n \in N$, then $a \geqslant 0$
if $a_n \leqslant b_n$ for all $n \in N$, then $a \leqslant b$
If there exists $c \in R$ for which $c \leqslant b_n$, for all $n \in N$, then $c \leqslant b$. Similarly, if $a_n < c$ for all $n \in N$, then $a \leqslant c$

Proof.

We will prove this by contraction; thus, let's assume $a < 0$. The idea is to produce a term in the sequence $a_n$ that is also less than zero. To do this, we consider the particular value $\epsilon = a$. The definition of convergence guarantee that we can find an $N$ such that $|a_n - a| < |a|$ for all $n \geqslant N$. In particular, this would mean that $|a_N - a| < |a|$, which implies $a_N < 0$. This contradicts our hypothesis that $a_N \geqslant 0$. We therefore conclude that $a \geqslant 0$
The Algebraic Limit Theorem ensures that the sequence $b_n - a_n$ converges to $b - a$. Because $b_n - a_n \geqslant 0$, we can apply previous proof to get that $b - a \geqslant 0$
Take $a_n = c$ (or $b_n = c$) for all $n \in N$, and apply previous proof.

Tags

[Understanding Analysis by Stephen Abbott] - [Chapter 2 Sequences and Series] - [2.3 The Algebraic and Order Limit Theorems]

Limits and Order