Advanced Calculus Ch.1

Chapter 1. Sequence and Series

Definition of Sequence

\[\text{A Sequence {$a_n$} of real numbers is a function from $\mathbb{N}$ to $\mathbb{R}$ mapping $n \mapsto a_n$}\]

Lemma

\[\begin{array}{l} \text{Let $a$ and $b$ be real numbers. then, }\\ a \leq b \Leftrightarrow \forall \epsilon > 0,a < b + \epsilon . \forall \epsilon > 0,|a - b| < \epsilon \Leftrightarrow a = b \end{array}\]

proof

$$ \begin{array}{l} \text{To prove } \Rightarrow, \; a \leq b < b + \epsilon \text{ since } \epsilon > 0 \\ \text{To prove } \Leftarrow, \; \text{we show its contradiction.} \\ \text{Suppose to contrary that } \forall \epsilon > 0, \, a < b + \epsilon \, \rightarrow \, a > b \\ \text{Let } \epsilon = a - b \text{ since } a > b. \text{ Then } a > a, \text{ which is contradiction.} \\\\ \text{By the before proof, } |a - b| \leq 0 \text{ (Replace a and b by $|a - b|$ and $0$ in that.)} \\ \text{Note that }|a - b| \geq 0.\,\, \text{thus } |a - b| = 0. \end{array} $$

Convergence of a Sequence

Definition

\[\{a_n\} \text{ is converge to real number $a$} \Leftrightarrow \forall \epsilon > 0, \exists N \in \mathbb{N} \text{ s.t } |a_n - a| < \epsilon\, \forall n \geq N.\]

Property

\(\text{The limit of a sequence ${a_n}$, if it exists, is unique}\)

proof

$$\begin{array}{l} \text{Let }\lim a_n = a, \lim a_n = b.\\ \text{Let $\epsilon > 0$ is given}\\ \text{There is $N_1, N_2 \in \mathbb{N}$ s.t } \forall n \geq N_1,\,|a_n - a| < \frac{\epsilon}{2} \text{ and } \,\forall n \geq N_2,|a_n - b| < \frac{\epsilon}{2}. \\ \text{Let } N = max\{N_1, N_2\}, \, |a - b| \leq |a - a_n| + |b - b_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \\ \text{Thus, } a = b. \end{array}$$

Divergence of a Sequence

Definition

\[\{a_n\} \text{ is diverges} \Leftrightarrow \text{$a_n$ does not converge to any real number.}\]

Bounded Sequence

Definition

\[\begin{array}{l} \{a_n\} \text{ is bounded above } \Leftrightarrow \text{range } \{a_n : n \in \mathbb{N}\} \text{ is bounded above.} \\ \{a_n\} \text{ is bounded below } \Leftrightarrow \text{range } \{a_n : n \in \mathbb{N}\} \text{ is bounded below.} \\ \{a_n\} \text{ is bounded } \Leftrightarrow \exists M > 0 \text{ s.t } \forall n \in \mathbb{N}, \,|a_n| \leq M \, \end{array}\]

Property

\[\begin{array}{l} \text{Every convergent sequence is bounded.} \end{array}\]

proof

$$\begin{array}{l} \text{Let $\{a_n\}$ be a convergent sequence to } a \in \mathbb{R}.\\ \text{Then there is $N \in \mathbb{N}$ s.t } |a_n| - |a| \leq |a_n - a| < 1 \,\forall n \geq N.\\ \text{Let }M = \max\{|a| + 1, |a_1|, \ldots, |a_{N-1}|\}. \text{ Then }\forall n \in \mathbb{N}, \,|a_n|\leq M. \end{array}$$

Basic Limit Calculation

\[\begin{array}{l} \text{Suppose that } a, b\in \mathbb{R}, \lim_{n\rightarrow \infty}a_n = a, \lim_{n\rightarrow \infty} b_n = b\\\\ \forall \alpha, \beta \in \mathbb{R}, \lim_{n\rightarrow \infty}(\alpha a_n + \beta b_n) = \alpha a + \beta b \\ \lim_{n\rightarrow \infty}a_nb_n = ab \\ \lim_{n\rightarrow \infty}\frac{a_n}{b_n} = \frac{a}{b} \text{ if }b \neq 0 \end{array}\]

proof

$$\begin{array}{l} \text{Let $\epsilon > 0$ be given. Since } \lim a_n = a,\, \lim b_n = b, \exists N_1, N_2 \in \mathbb{N} \\ \text{ s.t } \forall n \geq N_1,\, |a_n - a| < \frac{\epsilon}{2|\alpha| + 1}, \, \forall n \geq N_2,\, |b_n - b| < \frac{\epsilon}{2|\beta| + 1}.\\ \text{Take } N = \max\{N_1, N_2\}. \text{Then for all }n \geq N,\\ |(\alpha a_n + \beta b_n) - (\alpha a + \beta b)| \leq |\alpha||a_n - a| + |\beta||b_n - b| \leq \frac{\epsilon}{2|\alpha| + 1}+\frac{\epsilon}{2|\beta| + 1} < \epsilon \end{array}$$