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\begin{document}
\section{Sequences, Series and Limits}

\begin{defi}[Sequences and Convergence]
    Let $\metric$ be a metric space. A sequence is a mapping $\natn \rightarrow X$. We write $\seq{x}_{n \in \natn}$ or $\seq{x}$.

    The sequence $\seq{x}$ is said to be convergent to $x \in X$ if
    \[
        \forall \epsilon > 0 ~\exists N \in \natn ~\forall n \ge N: ~~d(x_n, x) < \epsilon
    \]
    $x$ is said to be the limit, and sequences that aren't convergent are called divergent.
\end{defi}

\begin{rem}
    On $\realn$ the metric is the Euclidian metric $|\cdot|$, therefore this new definition of convergence is merely a generalization of the old one.
\end{rem}

\begin{thm}
    Let $\seq{x}$ be a sequence in the metric space $\metric$ and $x \in X$. Then the following statements are equivalent:
    \begin{enumerate}[(i)]
        \item $\seq{x}$ converges to $x$
        \item $\forall \epsilon > 0$ $\oball(x)$ contains all but finitely many elements of the sequence (almost every (a.e.) element)
        \item $(d(x, x_n))$ is a null sequence
    \end{enumerate}
\end{thm}
\begin{proof}
    (ii) is merely a reformulation of (i), and $(ii) \iff (iii)$ follows from
    \begin{equation}
        d(x_n, x) = |d(x_n, x) - 0|
    \end{equation}
\end{proof}

\begin{thm}
    Let $\left(x^{(n)}\right) = (x_1^{(n)}, x_2^{(n)}, \cdots, x_d^{(n)}) \subset \realn^d$ and 
    \[
        x = (x_1, \cdots, x_d) \in \realn^d
    \]
    $\left(x^{(n)}\right)$ is said to converge to $x$ if and only if $x_i^{(n)}$ converges to $x_i$ for all $i$ in $\set{1, \cdots, d}$
\end{thm}
\begin{proof}
    For $y = (y_1, \cdots, y_d) \in \realn^d$ we have
    \begin{equation}
        \norm{y_i} < \norm{y} ~~\forall i \in \set{1, \cdots, d}
    \end{equation}
    If $\left(x^{(n)}\right)$ converges to $x$, then
    \begin{equation}
        \abs{x_i^{(n)} - x_i} \le \norm{x^{(n)} - x} \conv{} 0
    \end{equation}
    If $(x_i^{(n)})$ converges to $x_i ~~\forall i \in \set{1, \cdots d}$, then
    \begin{equation}
        \forall \epsilon > 0 ~\exists N \in \natn ~\forall n > N: ~~\abs{x_i^{(n)} - x_i} < \frac{\epsilon}{\sqrt{d}} ~~\forall i \in \set{1, \cdots d}
    \end{equation}
    Thus
    \begin{equation}
    \begin{split}
        \norm{x^{(n)} - x} &= \sqrt{(x_1^{(n)} - x_1)^2 + (x_2^{(n)} - x_2)^2 + \cdots + (x_d^{(n)} - x_d)^2} \\
        &\le \sqrt{\frac{\epsilon^2}{d} + \frac{\epsilon^2}{d} + \cdots + \frac{\epsilon}{2}} \\
        &= \epsilon
    \end{split}
    \end{equation}
    So $\left(x^{(n)}\right)$ converges to $x$.
\end{proof}

\begin{thm}
    Every convergent sequence has exactly one limit and is bounded.
\end{thm}
\begin{proof}
    Assume that $x, y$ are limits of $(x_n)$ with $x \ne y$. Then $d(x, y) > 0$. There exists $N_1, N_2 \in \natn$, such that
    \begin{subequations}
    \begin{align}
        d(x_n, x) &< \frac{d(x, y)}{2} ~~\forall n \ge N_1 \\
        d(x_n, x) &< \frac{d(x, y)}{2} ~~\forall n \ge N_2
    \end{align}
    \end{subequations}
    From this follows that
    \begin{equation}
        d(x, y) \le d(x, x_n) + d(x_n, y) < d(x, y) ~~\forall \max\set{N_1, N_2}
    \end{equation}
    which is a contradiction, thus sequences can have only one limit.

    Now if $\seq{x}$ converges to $x$, then 
    \begin{equation}
        \exists N \in \natn ~\forall n \ge N: ~~d(x_n, x) < 1
    \end{equation}
    Then 
    \begin{equation}
        d(x_n, x) \le \max\set{d(x_1, x), d(x_2, x), \cdots, d(x_{N-1}, x), 1}
    \end{equation}
\end{proof}

\begin{thm}
    Let $\normed$ be a normed space over $\field$. Let $\seq{x}, \seq{y} \subset V$ be sequences with limits $x, y \in V$ 
    and $\anyseqdef[\lambda]{\field}$ a sequence with limit $\lambda \in \field$. Then 
    \begin{align*}
        x_n + y_n \longrightarrow x + y && \lambda_n x_n \longrightarrow \lambda x
    \end{align*}
\end{thm}
\begin{proof}
    \reader
\end{proof}

\begin{defi}[Cauchy sequences and completeness]
    A sequence $\seq{x}$ in a metric space $\metric$ is called Cauchy sequence if 
    \[
        \forall \epsilon > 0 ~\exists N \in \natn: ~~d(x_n, x_m) < \epsilon ~~\forall m, n \ge N
    \]
    A metric space is complete if every Cauchy sequence converges. A complete normed space is called Banach space.
\end{defi}

\begin{eg}
    \item $(\realn, \abs{\cdot})$ and $(\cmpln, \abs{\cdot})$ are complete 
    \item $(\ratn, \abs{\cdot})$ is not complete
\end{eg}

\begin{thm}
    Every convering series is a Cauchy sequence
\end{thm}
\begin{proof}
    Let $\seq{x} \conv{} x$. This means that 
    \begin{equation}
        \forall \epsilon > 0 ~\epsilon N \in \natn: ~~d(x_n, x) < \frac{\epsilon}{2} ~~\forall n \ge N
    \end{equation}
    Then 
    \begin{equation}
        d(x_n, x_m) \le d(x_n, x) + d(x, x_m) < \epsilon ~~\forall m, n \ge N
    \end{equation}
\end{proof}

\begin{thm}
    $\realn^n$ with the Euclidian norm is complete.
\end{thm}
\begin{proof}
    Let $\left(x^{(n)}\right) \subset \realn^n$ be a Cauchy sequence. We know that 
    \begin{equation}
        \forall y \in \realn^n: ~~\abs{y_i} \le \norm{y} ~~\forall i \in \set{1, \cdots, n}
    \end{equation}
    We also know that $(x_i^{(n)})$ are Cauchy sequences because 
    \begin{equation}
        \abs{(x_i^{(n)} - x_i^{m})} \le \norm{x^{(n)} - x^{(m)}} ~~\forall i \in \set{1, \dots, n}
    \end{equation}
    Thus $x_i^{(n)} \conv{} x_i$ and therefore $\left(x^{(n)}\right) \conv{} x$.
\end{proof}

\begin{defi}[Series and (absolute) convergence]
    Let $\normed$ be a normed space and $\anyseqdef{V}$. The series 
    \[
        \series{k} x_k
    \]
    is the sequence of partial sums
    \[
        s_n = \series[n]{k} x_k
    \]
    If the series converges then $\series{k} x_k$ also denotes the limit. 
    The series is said to absolutely convergent if 
    \[
        \series{k} \norm{x_k} < \infty
    \]
\end{defi}

\begin{thm}
    In Banach spaces every absolutely convergent series is convergent.
\end{thm}
\begin{proof}
    Let $\normed$, $\anyseqdef{V}$ and require $\series{n} \normed{x_n} < \infty$. We need to show that $s_n = \series[n]{k} x_k$ is a Cauchy sequence.
    Let $\epsilon > 0$ and $t_n = \series[n]{k} \norm{x_k}$. $\seq{t}$ is convergent in $\realn$, and thus a Cauchy sequence.
    I.e. 
    \begin{equation}
        \exists N \in \natn: ~~|t_n - t| < \epsilon ~~\forall m, n \ge N
    \end{equation}
    For $n > m > N$:
    \begin{equation}
        \norm{s_n - s_m} = \norm{\sum_{k=m+1}^n x_k} \le \sum_{k=m+1}^n \norm{x_k} = t_n - t_m = |t_n - t_m| < \epsilon
    \end{equation}
\end{proof}

\begin{thm}
    Let $\normed$ be a Banach space, $\series{k} x_k$ absolutely convergent and let $\sigma: \natn \rightarrow \natn$ be a bijective mapping. Then 
    \[
        \series{k} x_k = \series{k} x_{\sigma(k)}
    \]
\end{thm}
\begin{proof}
    Analogous to $\Cref{259}$
\end{proof}
\end{document}