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  {\normalsize\textcolor{white!80}{<<lecture_number>>: Sequences and Convergence}}\par
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\begin{abstract}
\noindent These notes cover the fundamental theory of sequences in $\mathbb{R}$, including convergence criteria, boundedness, and the Bolzano--Weierstrass theorem.
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\vspace{12pt}

\section{Sequences in $\mathbb{R}$}

\begin{definitionbox}[Definition: Sequence]
A \textbf{sequence}\index{sequence} in $\mathbb{R}$ is a function $a\colon\mathbb{N}\to\mathbb{R}$. We write $(a_n)_{n=1}^{\infty}$ or simply $(a_n)$.
\end{definitionbox}

\marginnote{\footnotesize\textcolor{notegray}{Recall that $\mathbb{N} = \{1, 2, 3, \ldots\}$ in this course.}}

\begin{definitionbox}[Definition: Convergence]
A sequence $(a_n)$ \textbf{converges}\index{convergence} to a limit $L \in \mathbb{R}$ if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that
\[ n \geq N \implies |a_n - L| < \varepsilon. \]
We write $\lim_{n\to\infty} a_n = L$ or $a_n \to L$.
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\begin{examplebox}
\textbf{Example 1.1.} The sequence $a_n = 1/n$ converges to $0$. Given $\varepsilon > 0$, choose $N > 1/\varepsilon$ (by the Archimedean property). Then for $n \geq N$:
\[ |a_n - 0| = \frac{1}{n} \leq \frac{1}{N} < \varepsilon. \qedhere \]
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\section{Boundedness and Monotonicity}

\begin{theorem}[Convergent sequences are bounded]\index{bounded sequence}
\label{thm:conv-bounded}
If $(a_n) \to L$, then $(a_n)$ is bounded: there exists $M > 0$ such that $|a_n| \leq M$ for all $n$.
\end{theorem}

\begin{proof}
Take $\varepsilon = 1$. There exists $N$ such that $|a_n - L| < 1$ for all $n \geq N$. By the triangle inequality, $|a_n| < |L| + 1$ for $n \geq N$. Set
\[ M = \max\{|a_1|, |a_2|, \ldots, |a_{N-1}|, |L|+1\}. \]
Then $|a_n| \leq M$ for all $n \in \mathbb{N}$.
\end{proof}

\begin{remarkbox}
\textbf{Remark.} The converse is false: the sequence $a_n = (-1)^n$ is bounded but divergent.
\end{remarkbox}

\begin{theorem}[Monotone Convergence Theorem]\index{Monotone Convergence Theorem}
\label{thm:mct}
Every bounded monotone sequence in $\mathbb{R}$ converges.
\end{theorem}

\begin{proof}
Suppose $(a_n)$ is increasing and bounded above. Let $L = \sup\{a_n : n \in \mathbb{N}\}$, which exists by the completeness of $\mathbb{R}$. Given $\varepsilon > 0$, $L - \varepsilon$ is not an upper bound, so there exists $N$ with $a_N > L - \varepsilon$. Since $(a_n)$ is increasing, for $n \geq N$:
\[ L - \varepsilon < a_N \leq a_n \leq L < L + \varepsilon. \]
Thus $|a_n - L| < \varepsilon$ for all $n \geq N$.
\end{proof}

\section{The Bolzano--Weierstrass Theorem}

\begin{theorem}[Bolzano--Weierstrass]\index{Bolzano--Weierstrass theorem}
Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.
\end{theorem}

\begin{proof}[Proof sketch]
Let $(a_n)$ be bounded by $[-M, M]$. Apply the bisection method:
\begin{enumerate}[leftmargin=*]
  \item Set $I_1 = [-M, M]$.
  \item At step $k$, bisect $I_k$. At least one half contains infinitely many terms; call it $I_{k+1}$ and choose $a_{n_k} \in I_{k+1}$ with $n_k > n_{k-1}$.
  \item The nested intervals $I_k$ satisfy $|I_k| = 2M/2^{k-1} \to 0$, so by the Nested Intervals Theorem, $\bigcap_k I_k = \{L\}$ and $a_{n_k} \to L$.\qedhere
\end{enumerate}
\end{proof}

\marginnote{\footnotesize\textcolor{notegray}{This proof relies on the completeness of $\mathbb{R}$. It fails in $\mathbb{Q}$.}}

\begin{corollary}
A sequence in $\mathbb{R}$ converges if and only if it is a Cauchy sequence.
\end{corollary}

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