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\begin{fclemmastar}[]
  \glsnoundefn{eulc}{Euler's constant}{NA}%
  \glssymboldefn{gamma}%
  If \cloze[1]{$x \ge 1$}, then
  \[
    \sum_{n \le x} \frac{1}{n} = \log x +
    \mathcloze[1]{\gamma +
    \On \left( \frac{1}{x} \right)}
  ,\]
  where \cloze[1]{$\gamma \in \Rbb$}
  \cloze[2]{is Euler's constant,
  which is given by $\gamma = \lim_{N \to \infty}
  \sum_{k = 1}^N \frac{1}{k} - \log N$.}
\end{fclemmastar}

\begin{proof}
  Apply \nameref{lemma:partsum} with $a_n = 1$,
  $f(t) = \frac{1}{t}$, $y = \half$. Clearly $A(t) =
  \left\lfloor t \right\rfloor$. Then,
  \begin{align*}
    \sum_{n \le x} \frac{1}{n}
    &= \frac{\left\lfloor x \right\rfloor}{x}
    + \int_1^x \frac{\left\lfloor t
    \right\rfloor}{t^2} \dd t \\
    &= \frac{x + \On(1)}{x} + \int_1^x \frac{t -
    \{t\}}{t^2} \dd t \\
    &= 1 + \On \left( \frac{1}{x} \right) + \log x -
    \int_1^x \frac{\{t\}}{t^2} \dd t \\
    &= 1 + \log x - \int_1^\infty
    \frac{\{t\}}{t^2} \dd t + \On \left( \frac{1}{x} \right)
  \end{align*}
  The last equality is true since $\int_x^\infty
  \frac{\{t\}}{t^2} \dd t \le \int_x^\infty
  \frac{1}{t^2} \dd t = \frac{1}{x}$.

  Let $\gamma = 1 - \int_1^\infty
  \frac{\{t\}}{t^2} \dd t$. Then we have the
  asymptotic equation as desired.

  Taking $x \to \infty$ in the formula, we see
  that $\gamma$ is equal to the formula for
  Euler's constant, as desired.
\end{proof}

\begin{fclemmastar}[]
  For $x \ge 1$,
  \[
    \mathcloze{\sum_{p \le x} \frac{1}{p}}
    = \mathcloze{\int_1^x \frac{\pi(t)}{t^2} \dd t +
    \On(1)}
  .\]
\end{fclemmastar}

\begin{proof}
  Apply \cref{lemma:partsum} with $a_n =
  \indicator{\Pbb}(n)$ (where $\Pbb$ is the set
  of primes), $f(t) = \frac{1}{t}$, and $y = 1$.

  We get $A(t) = \pi(t)$, and then
  \begin{align*}
    \sum_{p \le x} \frac{1}{p}
    &= \frac{\pi(x)}{x} + \int_1^x
    \frac{\pi(t)}{t^2} \dd t \\
    &= \int_1^x \frac{\pi(t)}{t^2} \dd t + \On(1)
    \qedhere
  \end{align*}
\end{proof}

\subsection{Arithmetic Functions and Dirichlet
convolution}

\begin{fcdefnstar}[Arithmetic function]
  \glsnoundefn{af}{arithmetic function}{arithmetic
  functions}%
  An \emph{arithmetic function} is a function $f :
  \Nbb \to \Cbb$.
\end{fcdefnstar}

\begin{fcdefnstar}[Multiplicative]
  \glsadjdefn{mult}{multiplicative}{\gls{af}}%
  \glsadjdefn{cmult}{completely multiplicative}{\gls{af}}%
  An \gls{af} $f$ is \emph{multiplicative} if
  $f(1) = 1$ and $f(mn) = f(m) f(n)$ whenever $m,
  n \in \Nbb$ are coprime.

  Moreover, $f$ is \emph{completely
  multiplicative} if $f(mn) = f(m) f(n)$ for all
  $m, n$.
\end{fcdefnstar}

\begin{example*}
  \phantom{}
  \begin{itemize}
    \item $f(n) = n^s$ for $s \in \Cbb$ is
      \gls{cmult}.
    \item \glssymboldefn{mob}%
      M\"obius function
      \[
        \mu(n) = \begin{cases}
          1 & n = 1 \\
          (-1)^k & \text{if $n$ is a product of $k$
          distinct primes} \\
          0 & \text{$n$ is divisible by a square of
          a prime}
        \end{cases}
      \]
      This is \gls{mult}:
      \begin{itemize}
        \item If $\mu(mn) = 0$ and $m$, $n$ are
          coprime, then we must have had at least one of
          $\mu(m) = 0$ or $\mu(n) = 0$.
        \item If $\mu(mn) = 1$, then say $m$ is a
          product of $k$ distinct primes and $n$ is
          a product of $l$ distinct primes. Then
          $\mu(mn) = (-1)^{k + l} = (-1)^k (-1)^l =
          \mu(m)\mu(n)$.
        \item \glssymboldefn{tau}%
          $\tau(n) = \sum_{n = ab} 1$ (divisor
          function) and $\tau_k(n) = \sum_{n = n_1
          n_2 \cdots n_k} 1$ (generalised divisor
          function). $\tau$ and $\tau_k$ are
          \gls{mult}.
      \end{itemize}
  \end{itemize}
\end{example*}

On the space of \glspl{af}, we have operations:
\begin{align*}
  (f + g)(n)
  &= f(n) + g(n) \\
  (fg)(n)
  &= f(n) g(n) \\
  (f * g)(n)
  &= \text{Dirichlet convolution}
\end{align*}

\begin{fcdefnstar}[Dirichlet convolution]
  \glssymboldefn{*}%
  For $f, g : \Nbb \to \Cbb$, we define
  \[
    (f * g)(n) = \sum_{d \mid n} f(d) g \left(
    \frac{n}{d} \right)
  ,\]
  where $\sum_{d \mid n}$ means sum over the
  divisors of $n$.
\end{fcdefnstar}

\begin{fclemma}[]
  The space of \glspl{af} with \cloze{operations
  $+$, $\conv$} is a commutative ring.
\end{fclemma}

\begin{proof}
  Since \glspl{af} with $+$ form an abelian group,
  it suffices to show:
  \begin{enumerate}[(i)]
    \item $(f \conv g) \conv h = f \conv (g \conv h)$
    \item $f \conv g = g \conv f$
    \item $f \conv I = f$
    \item $f \conv (g + h) = f \conv g + f \conv h$
  \end{enumerate}
  Proofs:
  \begin{enumerate}[(i)]
    \item Follows from $(f \conv g)(n) = \sum_{n
      = ab} f(a) g(b)$.
    \item Follows from $(f \conv g)(n) = \sum_{n
      = ab} f(a) g(b)$.
    \item \glssymboldefn{I}%
      Take
      \[
        I(n) = \begin{cases}
          1 & n = 1 \\
          0 & n \neq 0
        \end{cases}
      .\]
      Then one can check that $f \conv I = f$.
    \item From definition.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{fclemmastar}[]
  The set of \glspl{af} $f$ with $f(1) \neq 0$
  form an abelian group with operation $\conv$.
\end{fclemmastar}

\begin{proof}
  Need $g$ such that $f \conv g = \dId$.
  \[
    (f \conv g)(1) = f(1) g(1) = 1
    \implies g(1) = \frac{1}{f(1)}
  .\]
  Assume $g(m)$ defined for $m < n$. We will
  defined $g(n)$.
  \begin{align*}
    (f \conv g)(n)
    &= g(n) f(1)
    + \sum_{\substack{d \mid n \\ d \neq 1}} f(d)
    g \left( \frac{n}{d} \right) \\
    \implies g(n)
    &= -\frac{1}{f(1)} \sum_{\substack{d \mid n \\
    d \neq 1}} f(d) g \ub{\left( \frac{n}{d} \right)}_{< n}
    \qedhere
  \end{align*}
\end{proof}
