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\subsubsection*{The Bolzano-Weierstrass Theorem}

\begin{theorem}
    If $x_n \in \RR$ and there exists $K$ such
    that $|x_n| \le K \,\,\forall\,\, n$, then we
    can find $n_1 < n_2 < n_3 < \cdots$ and $x \in
    \RR$ such that $x_{n_j} \to x$ as $j \to
    \infty$.
\end{theorem}
In other words every \emph{bounded} sequence has a
convergent subsequence.

\begin{remark*}
    We say nothing about uniqueness of $x$, for
    example $x_n = (-1)^n$, then $x_{2n + 1} \to
    -1$ and $x_{2n} \to 1$.
\end{remark*}

\begin{proof}
    Set $[a_1, b_1] = [-K, K]$. Let $c_n =
    \frac{a_n + b_n}{2}$ for all $n$. Consider the
    following possibilities:
    \begin{enumerate}[(1)]
        \item $x \in [a_1, c_1]$ for infinitely many
            values of $n$.
        \item $x_n \in [c_1, b_1]$ for infinitely
            many values of $n$.
    \end{enumerate}
    (1) and (2) could hold at the same time. But
    if (1) holds, we set $a_2 = a_1$ and $b_2 =
    c_1$. If (1) fails, we have that (2) must hold
    and we set $a_2 = c_1$ and $b_2 = b_1$.
    Proceed inductively to construct sequences
    $a_n, b_n$ such that $x_m \in [a_n, b_n]$ for
    infinitely many values of $m$.
    \[ a_{n - 1} \le a_n \le b_n \le b_{n - 1} \]
    \[ b_n - a_n = \frac{b_{n - 1} - a_{n - 1}}{2}
    \tag{$*$} \]
    (bisection method). Note that $a_n$ is an
    increasing sequence and bounded, and $b_n$ is
    a decreasing sequence and bounded, so by the
    Fundamental Axiom, $a_n \to a \in [a_1, b_1]$
    and $b_n \to b \in [a_1, b_1]$. Using ($*$),
    \[ b - a = \frac{b - a}{2} \implies a = b .\]
    Since $x_m \in [a_n, b_n]$ for infinitely many
    values of $m$, having chosen $n_j$ such that
    $x_{n_j} \in [a_j, b_j]$, there is $n_{j + 1} >
    n_j$ such that $x_{n_{j + 1}} \in [a_{j + 1},
    b_{j + 1}]$ (I have an ``unlimited supply''!)
    Since $a_j \le x_{n_j} \le b_j$, we have
    $x_{n_j} \to a$.
\end{proof}

\subsubsection*{Cauchy Sequences}

\begin{definition*}[Cauchy Sequence]
    $a_n \in \RR$ is called a \emph{Cauchy
    sequence} if given $\eps > 0$, $\exists\,\, N > 0$
    such that $|a_n - a_m| < \eps \,\,\forall\,\,
    n, m \le N$. (Note: $N = N(\eps)$.)
\end{definition*}

\begin{lemma}
    A convergent sequence is a Cauchy sequence.
\end{lemma}

\begin{proof}
    If $a_n \to a$, given $\eps > 0$, $\exists\,\, N$
    such that $\forall\,\, n \ge N$, $|a_n - a| <
    \eps$. Take $m, n \ge N$, then
    \[ |a_n - a_m| \le |a_n - a| + |a_m - a| <
    2\eps .\]
\end{proof}

\begin{theorem}
    Every Cauchy sequence is convergent.
\end{theorem}

\begin{proof}
    First we note that if $a_n$ is Cauchy, then it
    is \emph{bounded}. Take $\eps = 1$, $N =
    N(1)$ in the Cauchy property, then
    \[ |a_n - a_m| < 1, \quad \forall\,\, n, m \ge
    N(1) \]
    \[ |a_m| \le |a_m - a_N| + |a_N| < 1 + |a_N|
    \quad \forall\,\, m \ge N .\]
    Let $K = \max\{1 + |a_N|, |a_n|, n = 1, 2,
    \dots, N - 1\}$. Then $|a_n| \le K
    \,\,\forall\,\, n$. So by the
    Bolzano-Weierstrass theorem, $a_{n_j \to a}$.
    \myskip
    Claim: $a_n \to a$. \\
    We now prove the claim: given $\eps > 0$,
    $\exists\,\, j_0$ such that $\forall\,\, j \ge
    j_0$
    \[ |a_{n_j} - a| < \eps .\]
    Also, $\exists\,\, N(\eps)$ such that $|a_m - a_n|
    < \eps \,\,\forall\,\, m, n \ge N(\eps)$. Take
    $j$ such that $n_j \ge \max\{N(\eps),
    n_{j_0}\}$. Then if $n \ge N(\eps)$
    \[ |a_n - a| \le \ub{|a_n - a_{n_j}|}_{< \eps}
    + \ub{|a_{n_j} - a|}_{< \eps} < 2\eps .\]
\end{proof}

\noindent
Summary: in $\RR$ a sequence is convergent is and
only if it is Cauchy. \\
``old fashioned name'': the ``general principle of
convergence''. \\
Useful property: since we do not need to know what
the limit is.

\subsubsection*{Series}

\begin{definition*}
    $a_n \in \RR, \CC$. We say that $\sum_{j =
    1}^\infty a_j$ converges to $S$ if the
    sequence of partial sums
    \[ S_N = \sum_{j = 1}^N a_j \to S \]
    as $N \to \infty$. We write
    \[ \sum_{j = 1}^\infty a_j = S .\]
    If $S_N$ does not converge, we say that
    $\sum_{j = 1}^\infty a_j$ \emph{diverges}.
\end{definition*}

\begin{remark*}
    Nay problem in series is really a problem
    about the sequence of partial sums.
\end{remark*}

\begin{lemma}
    \begin{enumerate}[(i)]
        \item If $\sum_{j = 1}^\infty a_j$ and
            $\sum_{j = 1}^\infty b_j$ converge,
            then so does
            \[ \sum_{j = 1}^\infty (\lambda a_j +
            \mu b_j) \]
            where $\lambda, \mu \in \CC$.
        \item Suppose $\exists\,\, N$ such that $a_j =
            b_j \,\,\forall\,\, j \ge N$ then
            either $\sum_{j = 1}^\infty a_j$ and
            $\sum_{j = 1}^\infty b_j$ both
            converge or they both diverge (initial
            terms do not matter).
    \end{enumerate}
\end{lemma}

\begin{proof}
    \begin{enumerate}[(i)]
        \item Exercise
        \item For $n \ge N$,
            \[ s_n = \sum_{j = 1}^n a_j = \sum_{j
            = 1}^{N - 1} a_j + \sum_{j = N}^n a_j \]
            \[ d_n = \sum_{j = 1}^n b_j = \sum_{j
            = 1}^{N - 1} b_j + \sum_{j = N}^n b_j \]
            \[ \implies s_n - d_n = \sum_{j =
            1}^{N - 1} a_j - \sum_{j = 1}^{N - 1}
            b_j = \text{constant} \]
            So $s_n$ converges if and only if
            $d_n$ does.
    \end{enumerate}
\end{proof}
