Convergence of sequences in $\mathbb{R}$
Definition. We say that $a_{n} \xi_{n=1}^{\infty}(\subseteq X), X$ : topological space is a sequence if it is a function
$$
\begin{aligned}
\left\{a_{n}\right\}_{n=1}^{\infty}: \mathbb{N} & \longrightarrow X \\
n \longmapsto & a_{n}
\end{aligned}
$$
In advanced Calculus (or Introduction to Real Analysis), $X$ is mostly $\mathbb{R}^k, k\in \mathbb{N}$.
Definition. We say a sequence $a_{n} \xi_{n=1}^{\infty}(\subseteq \mathbb{R})$ converges to $\alpha \in \mathbb{R}$ if for each $\varepsilon>0$, there exists $N_{\varepsilon} \in \mathbb{N}$ such that for any $n \geq N_{\varepsilon}, n \in \mathbb{N}, \quad\left|a_{n}-\alpha\right|<\varepsilon$
Denotes $\lim _{n \rightarrow \infty} a_{n}=\alpha$.
Otherwise, we say the sequence {$\left. a_{n}\right\}$ diverges.
Remark.
Step (0) For “each” $\varepsilon>0, (\rightarrow \varepsilon$ is fixed).
Step (1) there exists (Correspanding) $N_{\varepsilon} \in \mathbb{N}$.
Step (2) for any $n \geq N_\varepsilon, n \in \mathbb{N}$, ( $\rightarrow n$ is fixed, but arbitary).
Step (3) $\left|a_{n}-\alpha\right|<\varepsilon$.
ex) $a_{n}=\frac{1}{n}$, for each $n \in \mathbb{N}$
Claim: $\lim _{n \rightarrow \infty} \frac{1}{n}=0$
$:$ Fix any $\varepsilon>0$
By Archimedean property, there $N_{\varepsilon} \in \mathbb{N}$ $(a=\varepsilon, b=1)$
such that $N_{\varepsilon} \cdot \varepsilon>1 \Leftrightarrow \frac{1}{N_{\varepsilon}}<\varepsilon$.
Then for any $n \geq N_{\varepsilon} \Leftrightarrow \frac{1}{n} \leq \frac{1}{N_{\varepsilon}}$,
$$
\left|a_{n}-x\right|=\left|\frac{1}{n}-0\right|=\frac{1}{n} \leq \frac{1}{N_{\varepsilon}}<\varepsilon.
$$
i.e., $|a_n – \alpha | <\varepsilon$.
Hence $\lim _{n \rightarrow \infty} a_{n}=0$.
Counter- example) $b_{n}=1-(-1)^{n}$ for each n. Then $(b_n)$ diverges.
Observe that $\left|b_{m}-b_{m+1}\right|=2>0$ for any $m \in \mathbb{N}$.
Choose $\varepsilon>0$ satisfying $\varepsilon < 2$.
If $(b_n)$ converges to some $\beta \in \mathbb{R}$, then
$$
\begin{aligned}
\left|b_{n}-\beta\right| & =\left|b_{n}-b_{n+1}+b_{n+1}-\beta\right| \\
& \leq \frac{b_{n}-b_{n+1}|+| b_{n+1}-\beta \mid}{\frac{21}{2}} \\
& \geq 2
\end{aligned}
$$
but also
$<\varepsilon$ : impossible, for any $n \in \mathbb{N}$.
Thus there is no $N_{\varepsilon} \in \mathbb{N}$ Satisfying for any $n \geq N_{\varepsilon},\left|b_{n}-\beta\right|<\varepsilon$
Remark. (Property of the absolute value)
$$
|x|=\left\{\begin{array}{cl}
x & \text { if } x \geq 0 \\
-x & \text { if } x<0
\end{array}\right.
$$
(a) $|x|=|-x|=\sqrt{x^{2}}$
(b) $|x y|=|x| \cdot|y|$
(c) For each $r>0,|x|<r$ { iff }$-r<x<r$
(d) $-|x| \leq x \leq|x|$
(e) $|x+y| \leq|x|+|y|$
(f) $|x|>\delta$ iff $x>\delta$ or $x<- \delta$.
Definition. A sequence \{a_n $\}_{u=1}^{\infty}$ in $\mathbb{R}$ is bounded if there exists $M>0$ arch that for any $n \in \mathbb{N}$, $\left|a_{n}\right| \leq M$.
Remark.
(Wrong) A sequence $(a_n)$ in $\mathbb{R}$ is bounded if for any $n \in \mathbb{N}$, there exists $M>0$ such tot
$$
\left|a_{n}\right| \leq M=\left|a_{n}\right|.
$$
Note. Boundness is NOT a topolgyical property.
Proposition. If $\left\{a_{n}\right\}_{n=1}^{\infty}$ in $\mathbb{R}$ is convergent, then $\left\{a_{n}\right\}$ is bounded.
Proof. Let $\alpha=\lim _{n \rightarrow \infty} a_{n} \in \mathbb{R} \operatorname{coog} a_{n s}$ is convergent
Take $\varepsilon=1$
Then there exists $N_{1} \in \mathbb{N}$ such such that for any $n \geq N_{1}, n \in \mathbb{N},\left|a_{n}-\alpha\right|<\varepsilon=1$,
$\Leftrightarrow \alpha-1 < a_n < \alpha+1$.
Then we can choose $M=$ max $\left.|\alpha+1|,\left|a_{1}\right|, \ldots,\left|a_{N-1}\right|\right\}$ .
Then for any $n \in \mathbb{N},\left|a_{n}\right| \leq M$.
Question. What about the converse?
i.e if if $(a_{n})_{n=1}^{\infty} \in \mathbb{R}$ is bounded then is it convergent?
Answer. In general, No, (counter-ex) $b_{n}=\left(1-(-1)^{n}\right.$
However, “partially” true (Bolzano-Weierstrass theorem).
Next: Limit theorem Let $\operatorname{lm} a_{n}=\alpha, \lim b_{n}=\beta$.
(a) $\operatorname{lm}\left(a_{n} \pm b_{n}\right)=\alpha \pm \beta$,
(b) $\lim \left(a_{n} b_{n}\right)=\alpha \beta$,
(c) $\operatorname{lm}\left(k a_{n}\right)=k \alpha$, for ans $k \in \mathbb{R}$,
(d) $\operatorname{lm}\left(\frac{a_{n}}{b_{n}}\right)=\frac{\alpha}{\beta}$ provided $b_{n}, \beta \neq 0$.