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\begin{proof}[Proof of (2)]
  \begin{align*}
    \prod_{p \le x} \left( 1 - \frac{1}{p} \right)^{-1}
    &= \prod_{p \le x} \left( 1 + \frac{1}{p} +
    \frac{1}{p^2} + \cdots \right) \\
    &= \sum_{k_1, \ldots, k_r \ge 0} (p_1^{k_1}
    \cdots p_r^{k_r})^{-1}
  \end{align*}
  Every integer $1 \le n \le x$ is a product of
  \glspl{prime} $\le x$, so
  \[ \prod_{p \le x} \left( 1 - \frac{1}{p}
  \right)^{-1} \ge \sum_{1 \le n \le x}
  \frac{1}{n} \to \infty \]
  as $x \to \infty$ (harmonic series).
\end{proof}

\begin{flashcard}[riemann-zeta-func-defn]
\begin{definition}[Riemann $\zeta$]
  \glssymboldefn{zeta_fn}{$\zeta$}{$\zeta$}
  \cloze{
  The \emph{Riemann $\zeta$-function} is
  \[ \zeta(s) = \sum_{n = 1}^\infty n^{-s} .\]
  Convention: $s \in \CC$. This defines $\zeta(s)$
  whenever this series converges.
  }
\end{definition}
\end{flashcard}
\vspace{-1em}

$\zetafn(s)$
studied by Euler for $s \in \RR$ by Riemann for
$s \in \CC$ $\to$ complex analysis.

\begin{proposition}
  If $s \in \CC$, $\Re(s) > 1$, then $\zetafn(s)$
  converges absolutely.
\end{proposition}

\begin{notation*}
  \glssymboldefn{sigmareal}{$\sigma$}{$\sigma$}
  Notation: $s = \sigmareal + it$, $\sigmareal + it \in
  \RR$.
\end{notation*}

\begin{proof}
  \[ n^{-s} = \exp(-s\log n) = \exp(-(\sigmareal +
  it)\log n) \]
  \[ \implies |n^{-s}| = \exp(-\sigmareal \log n) =
  n^{-\sigmareal} \]
  So $\sum_{n = 1}^\infty |n^{-s}| = \sum_{n =
  1}^\infty n^{-\sigmareal}$. This converges if and
  only if $\sigmareal > 1$.
\end{proof}

Same arugument shows that $\zetafn(s)$ converges
uniformly in $\{s \in \CC \st \sigmareal > 1 +
\delta\}$, for any $\delta > 0$. A uniform limit
of holomorphic functions is holomorphic, so
$\zetafn(s)$ is holomorphic in $\{s \in \CC \st
\sigmareal > 1\}$.

\begin{flashcard}[zeta-prime-product-thm]
\begin{theorem}
  If $s \in \CC$, $\sigmareal > 1$, then
  \[ \zetafn(s) = \cloze{\prod_{\text{$p$
  \gls{prime}}} (1 -
  p^{-s})^{-1}} .\]
  More precisely
  \[ \cloze{\lim_{x \to \infty} \prod_{p \le x} (1 -
  p^{-s})^{-1}} = \zetafn(s) \]
  and this limit is non-zero.
\end{theorem}
\end{flashcard}

\begin{proof}
  Arguing informally, we have
  \[ \prod_{p} (1 - p^{-s})^{-1} = \prod_p (1 +
  p^{-s} + p^{-2s} + p^{-3s} + \cdots) = \sum_{n =
  1}^\infty n^{-s} .\]
  By \nameref{FTA}.

  Arguing rigorously,
  \[ \prod_{p \le x} (1 - p^{-s})^{-1} =
  \sum_{k_1, \ldots, k_r} \ge 0 (p_1^{k_1} \cdots
  p_r^{k_r})^{-s} \]
  where $p_1, \ldots, p_r$ are the \glspl{prime}
  $\le x$. \nameref{FTA} implies if $n \in \NN$,
  then $n^{-s}$ apprears at most once in
  $\sum_{k_1, \ldots, k_r \ge 0} (p_1^{k_1} \cdots
  p_r^{k_r})^{-s}$, and exactly once if $n \le x$.
  So
  \[ \left|\prod_{p \le x} (1 - p^{-s})^{-1} -
  \zetafn(s)\right| \le \sum_{n > x} n^{-\sigmareal} \to
  0 \]
  as $x \to \infty$. So
  \[ \lim_{x \to \infty} p \le x (1 - p^{-s})^{-1}
  = \zetafn(s) .\]
  To show $\zetafn(s) \neq 0$, consider
  \[ \prod_{p \le x} (1 - p^{-s}) \zetafn(s) =
  \prod_{p > x}(1 - p^{-s})^{-1} = 1 + \sum_{n \in
  S_x} n^{-s} \]
  where
  \[ S_x = \{n \in \NN \st \text{all prime factors
  $p \divides n$ satisfy $p > x$}\} \subset \{n
  \in \NN \st n > x\} .\]
  Then
  \[ \left|\prod_{p \le x} (1 - p^{-s})
    \zetafn(s)\right| \ge
  1 - \sum_{n > x} n^{-\sigmareal} .\]
  Since $\sigmareal > 1$, $\sum_{n > x}
  n^{-\sigmareal} \to 0$ as $X \to \infty$, so we
  can choose $x$ such that
  \[ 1 - \sum_{n > x} n^{-\sigmareal} > 0. \]
  Then we deduce that
  \[ \left| \prod_{p \le x} (1 - p^{-s})
  \zetafn(s) \right| \neq 0 \implies \zetafn(s) \neq
  0. \qedhere \]
\end{proof}

\subsubsection*{Non-examinable discussion of
$\zeta(s)$}

\begin{itemize}
  \item Meromorphic continuation: $\zetafn(s)$
    admits a unique function on $\CC$, with a
    simple polt at $s = 1$, and no other poles.
  \item
    \glssymboldefn{Gamma_fn}{$\Gamma$}{$\Gamma$}
    \glssymboldefn{xi_fn}{$\xi$}{$\xi$}
    Functional equation: we define $\xi(s) =
    \pi^{-s/2} \Gamma(s/2) \zeta(s)$, where
    $\Gamma(s)$ is the \emph{Gamma function}, a
    meromorphic function in $\CC$ defined for
    $\sigmareal > 0$ by the integral
    \[ \Gamma(s) = \int_{y = 0}^\infty e^{-y} y^s
    \frac{\dd y}{y} .\]
    Then $\xi(s) = \xi(1 - s)$.
  \item Trivial zeroes: $\xifn(s)$ is meromorphic
    with simple poles at $s = 0$, $s = 1$ and no
    other poles. $\Gammafn(s)$ has simple poles at
    $s = 0, -1, -2, \ldots$ and no other poles.
    $\Gammafn(s/2)$ has simple poles at $s = 0, -2,
    -4, \ldots$. Since $\xifn$ is holomorphic at $s
    = -2, -4, -6, \ldots$ but $\Gammafn(s/2)$ has a
    pole, $\Gammafn(s)$ must vanish, whenever $s$ is
    a negative even integer (these are the trivial
    zeroes). Picture of $\zetafn(s)$:
    \begin{center}
      \includegraphics[width=0.6\linewidth]{images/604e3340e5f24ed5.png}
    \end{center}
  \item Critical strip: this is the region $\{s
    \in \CC \st \sigmareal \in [0, 1]\}$. All
    non-trivial zeroes of $\zetafn(s)$ lie in the
    critical strip.

    Fact: their location is closely related to the
    distribution of \glspl{prime}. For example,
    the ``hard part'' in the proof of
    \nameref{pnt} (\cref{pnt}) is the
    non-existence of zeroes of $\zetafn(s)$ with
    $\sigmareal = 1$.

    \begin{conjecture}[Riemann Hypothesis]
      \vspace{0.5em}
      If $s \in \CC$ is a non-trivial zero of
      $\zetafn(s)$, then $\sigmareal = \half$.
    \end{conjecture}
    \vspace{-1em}

    As stated in the first lecture, this is
    equivalent to the bound
    \[ |\pc(x) - \li(x)| \le \sqrt{x} \log x \]
    for any $x \ge 3$. Recall \nameref{pnt} says
    \[ \left| \frac{\pc(x)}{\li(x)} - 1 \right|
    \to 0 \]
    as $x \to \infty$.
\end{itemize}
\textbf{This is now the end of the non-examinable
content.}

\begin{flashcard}[dirichlet-series-defn]
\begin{definition}[Dirichlet series]
  \glsnoundefn{d_series}{Dirichlet series}{N/A}
  \cloze{A Dirichlet series is one of the form
  \[ \sum_{n = 1}^\infty a_n n^{-s} \qquad a_n \in
  \CC \]}
\end{definition}
\end{flashcard}
\vspace{-1em}

\begin{example*}
  If $a_n = 1 ~\forall n \in \NN$, this is just
  $\zetafn(s)$.
  
  If $N \in \NN$ is odd, then the \gls{d_series}
  \[ \sum_{n = 1}^\infty \legendre{n}{N} n^{-s} \]
  plays a rule in the proof of \cref{pnt_modulo}
  analogous to the role of $\zetafn(s)$ in the
  proof of \cref{pnt}.
\end{example*}

\begin{remark*}
  If $A, B > 0$ and $|a_n| \le An^B$ for all $n
  \ge 1$, then $\sum_{n = 1}^\infty a_n n^{-s}$
  converges absolutely whe $\sigma > 1 + B$.
\end{remark*}