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  So for $x \ge 2$ and a suitable large enough
  constant $c'$, we have
  \[
    \sum_{p \le x} \le (\log 4) x + c'(\log x)^2
  .\]
  Hence
  \begin{align*}
    \sum_{\frac{x}{(\log x)^2} < p \le x} \log p
    &\le (\log 4) x + c'(\log x)^2 \\
    \implies \log \frac{x}{(\log x)^2} \left(
    \pi(x) - \pi \left( \frac{x}{(\log x)^2} \right) \right)
    &\le (\log 4) x + c'(\log x)^2 \\
    \implies \pi(x)
    &\le \frac{x}{(\log x)^2} + (\log 4)
    \frac{x}{\log \frac{x}{(\log x)^2}} + c'
    \frac{(\log x)^2}{\log \frac{x}{(\log x)^2}}
    \\
    &\le (\log 4 + \eps) \frac{x}{\log x}
  \end{align*}
  for any $\eps$, as long as $x \ge x(\eps)$.

  Take $\eps = 1$. Choose $c > 0$ large enough.
\end{proof}

\subsection{Asymptotic Notation}

\begin{fcdefn}[Big $O$ and little $o$ notation]
  \glssymboldefn{o}%
  Let $f, g, h : S \to \Cbb$, $S \subseteq \Cbb$.

  Write $f(x) = O(g(x))$ if there is $c > 0$ such
  that $|f(x)| \le c|g(x)|$ for all $x \in S$.

  Write $f(x) = o(g(x))$ if for any $\eps > 0$
  there is $x_\eps > 0$ such that $|f(x)| \le
  \eps|g(x)|$ for $x \in S$, $|x| \ge x_\eps$.

  Write $f(x) = g(x) + O(h(x))$ if $f(x) - g(x) =
  O(h(x))$ and write $f(x) = g(x) + o(h(x))$ if
  $f(x) - g(x) = o(h(x))$.
\end{fcdefn}

\begin{fcdefn}[Vinogradov notation]
  \glssymboldefn{vn}%
  Let $f, g, h : S \to \Cbb$, $S \subseteq \Cbb$.

  Write $f(x) \ll g(x)$ or $g(x) \gg f(x)$ if
  $f(x) = \On(g(x))$.
\end{fcdefn}

\begin{example*}
  \phantom{}
  \begin{itemize}
    \item $(\log x)^100 \vll \exp(\sqrt{\log x})
      \vll x^{\frac{1}{100}}$ ($x \le 1$), since
      $\lim_{x \to \infty} \frac{(\log
      x)^{100}}{\exp(\sqrt{\log x})} = 0$,
      $\lim_{x \to \infty} \frac{\exp(\sqrt{\log
      x})}{x^{\frac{1}{100}}} = 0$.
    \item $100x + 100 \vll x \vll \frac{x}{100}$
      (for $x \ge 1$).
    \item $e^x = 1 + x + \On(x^2)$ for $x \in
      [-10, 10]$, since $e^x = 1 + x + \sum_{n =
      2}^{\infty} \frac{x^n}{n!}$.
    \item $\left\lfloor x \right\rfloor = x +
      \On(1)$ for $x \in \Rbb$ (since $\left\lfloor
      x\right\rfloor \in (x - 1, x]$).
    \item $\frac{x + 1}{x} = 1 + \on(1)$ (for $x \ge
      1$).
  \end{itemize}
\end{example*}

\begin{lemma*}
  Let $f, g, h, u : S \to \Cbb$.
  \begin{enumerate}[(i)]
    \item If $f(x) = \On(g(x))$ and $g(x) =
      \On(h(x))$, then $f(x) = \On(h(x))$
      (transitivity).
    \item If $f(x) = \On(h(x))$ and $g(x) =
      \On(u(x))$, then $f(x) + g(x) = \On(|h(x)| +
      |u(x)|)$.
    \item If $f(x) = \On(h(x))$ and $g(x) =
      \On(u(x))$, then $f(x) g(x) = \On(h(x) u(x))$.
  \end{enumerate}
\end{lemma*}

\begin{proof}
  Follows from the definition in a straightforward
  way. Example:
  \begin{enumerate}[(i)]
    \setcounter{enumi}{2}
    \item $|f(x)| \le c_1 |h(x)|$, $|g(x)| \le
      c_2|u(x)|$. Then $|f(x)g(x)| \le c_1
      c_2|h(x)u(x)|$, so $f(x)g(x) = \On(h(x)u(x))$.
      \qedhere
  \end{enumerate}
\end{proof}

\subsection{Partial Summation}

\begin{fclemma}[Partial Summation]
  \label{lemma:partsum}
  Assuming:
  - $(a_n)_{n \in \Nbb}$ are complex numbers
  - $x \ge y \ge 0$
  - $f : [y, x] \to \Cbb$ is continuously
  differentiable
  Then:
  \[
    \sum_{y < n \le x} a_n f(n) = A(x) f(x) - A(y)
    f(y) - \int_y^x A(t) f'(t) \dd t
  ,\]
  where for $t \ge 1$, we define
  \[
    A(t) = \sum_{n \le t} a_n = \sum_{n =
    1}^{\lfloor t \rfloor} a_n
  .\]
\end{fclemma}

\begin{proof}
  It suffices to prove the $y = 0$ case, since
  then
  \[
    \sum_{y < n \le x} a_n f(n) = \sum_{0 < n \le
    x} a_n f(n) - \sum_{0 < n \le y} a_n f(n)
  .\]
  Suppose $y = 0$. By the fundamental theorem of
  calculus,
  \[
    f(n) = f(x) - \int_n^x f'(t) \dd t
    = \int_0^x f'(t) \indicator{[n, x]} (t) \dd t
  .\]
  Summing over $n \le x$, we get
  \begin{align*}
    \sum_{n \le x} a_n f(n)
    &= A(x) f(x) - \int_0^x f'(t) \left( \sum_{n
    \le x} \indicator{[n, x]}(t) a_n \right) \dd t
    \\
    &= A(x) f(x) - \int_0^x f'(t) A(t) \dd t
    \qedhere
  \end{align*}
\end{proof}
