py.
\end{equation*}
Note that these pairs are precisely those where
\begin{equation*}
1\leq x\leq (p-1)/2 \mbox{and} \ \ 1\leq y\leq qx/p.
\end{equation*}
For each fixed value of $x$ with $1\leq x\leq (p-1)/2$, there are
$[qx/p]$ integers satisfying $1\leq y\leq qx/p$. Consequently, the
total number of pairs with are
\begin{equation*}
1\leq x\leq (p-1)/2, \ \ 1\leq y\leq qx/p, \mbox{and} \ \ qx>py
\end{equation*}
is
\begin{equation*}
\sum_{i=1}^{(p-1)/2}[qi/p].
\end{equation*}
\par Consider now the pair of integers $(x,y)$ with
\begin{equation*}
1\leq x\leq (p-1)/2, \ \ 1\leq y\leq (q-1)/2, \mbox{and} \ \ qx__\log P(x)-\frac{1}{2}
\end{equation*}
And thus $S(x)$ diverges as $x$ approaches infinity.
\end{proof}
\index{Abel Summation Formula}
\begin{theorem}[Abel's Summation Formula]\label{1}
For any arithmetic function $f(n)$, we let
\begin{equation*}
A(x)=\sum_{n\leq x}f(n)
\end{equation*}
where $A(x)=0$ for $x<1$. Assume also that $g$ has a continuous
derivative on the interval $[y,x]$, where $0 0$, we have
\begin{equation*}
0 \leq \frac{\psi(x)}{x}-\frac{\theta(x)}{x}\leq \frac{(\log
x)^2}{2\sqrt{x}\log 2}.
\end{equation*}
\end{theorem}
\begin{proof}
From Remark 4, it is easy to see that
\begin{equation*}
0\leq \psi(x)-\theta(x)=\theta(x^{1/2})+
\theta(x^{1/3})+...\theta(x^{1/m})
\end{equation*}
where $m\leq log_2x$. Moreover, we have that $\theta(x)\leq x\log
x$. The result will follow after proving the inequality in Exercise
2.
\end{proof}
\textbf{Exercises}
\begin{enumerate}
\item{Show that \begin{equation*}
\psi(x)=\theta(x)+\theta(x^{1/2})+
\theta(x^{1/3})+...\theta(x^{1/m})
\end{equation*}
where $m\leq log_2x$.}
\item{Show that $0\leq \psi(x)-\theta(x)\leq (\log_2(x))\sqrt{x}\log\sqrt{x}$ and thus the result of Theorem 86 follows.}
\item{Show that the following two relations are equivalent
\begin{equation*}
\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{\log^2x}\right)
\end{equation*}
\begin{equation*}
\theta(x)=x+O\left(\frac{x}{\log x}\right)
\end{equation*}}
\end{enumerate}
\section{Getting Closer to the Proof of the Prime Number Theorem}
We know prove a theorem that is related to the defined functions
above. Keep in mind that the prime number theorem is given as
follows:
\begin{equation*}
\lim_{x \rightarrow \infty} \frac{\pi(x)logx}{x}=1.
\end{equation*}
We now state equivalent forms of the prime number theorem.
\begin{theorem}
The following relations are equivalent
\begin{equation}\label{3}
\lim_{x\rightarrow \infty}\frac{\pi(x)\log x}{x}=1
\end{equation}
\begin{equation}\label{4}
\lim_{x\rightarrow \infty} \frac{\theta(x)}{x}= 1
\end{equation}
\begin{equation}\label{5}
\lim_{x\rightarrow \infty} \frac{\psi(x)}{x}= 1.
\end{equation}
\end{theorem}
\begin{proof}
We have proved in Theorem 86 that $(\ref{4})$ and $(\ref{5})$ are
equivalent, so if we show that $(\ref{3})$ and $(\ref{4})$ are
equivalent, the proof will follow. Notice that using the integral
representations of the functions in Theorem 85, we obtain
\begin{equation*}
\frac{\theta(x)}{x}=\frac{\pi(x)\log x}{x}-\frac{1}{x}\int_
{2}^{x}\frac{\pi(t)}{t}dt
\end{equation*}
and
\begin{equation*}
\frac{\pi(x)\log x}{x}=\frac{\theta(x)}{x}+\frac{\log x}{x}\int_{2}^x\frac{\theta(t)}{t\log^2t}dt.
\end{equation*}
Now to prove that (\ref{3}) implies $(\ref{4})$, we need to prove
that
\begin{equation*}
\lim_{x\rightarrow \infty}\frac{1}{x}\int_
{2}^{x}\frac{\pi(t)}{t}dt=0.
\end{equation*}
Notice also that $(\ref{3})$ implies that
$\frac{\pi(t)}{t}=O\left(\frac{1}{\log t}\right)$ for $t\geq 2$ and
thus we have
\begin{equation*}
\frac{1}{x}\int_
{2}^{x}\frac{\pi(t)}{t}dt=O\left(\frac{1}{x}\int_2^x\frac{dt}{\log
t}\right)
\end{equation*}
Now once you show that (Exercise 1)
\begin{equation*}
\int_2^x\frac{dt}{\log t}\leq \frac{\sqrt{x}}{\log
2}+\frac{x-\sqrt{x}}{\log \sqrt{x}},
\end{equation*}
then $(\ref{3})$ implies $(\ref{4})$ will follow. We still need to
show that $(\ref{4})$ implies $(\ref{3})$ and thus we have to show
that
\begin{equation*}
\lim_{x\rightarrow \infty}\frac{\log x}{x}\int_{2}^x
\frac{\theta(t)dt}{t\log^2t}=0.
\end{equation*}
Notice that $\theta(x)=O(x)$ and hence
\begin{equation*}
\frac{\log x}{x}\int_{2}^x
\frac{\theta(t)dt}{t\log^2t}=O\left(\frac{\log
x}{x}\int_2^x\frac{dt}{\log^2t}\right).
\end{equation*}
Now once again we show that (Exercise 2)
\begin{equation*}
\int_2^x\frac{dt}{\log^2t}\leq
\frac{\sqrt{x}}{\log^22}+\frac{x-\sqrt{x}}{\log^2\sqrt{x}}
\end{equation*}
then $(\ref{4})$ implies $(\ref{3})$ will follow.
\end{proof}
\begin{theorem}
Define
\begin{equation*}
l_1=\liminf_{x\rightarrow \infty}\frac{\pi(x)}{x/log x}, \ \ \ \ \
L_1=\limsup_{x\rightarrow \infty}\frac{\pi(x)}{x/log x},
\end{equation*}
\begin{equation*}
l_2=\liminf_{x\rightarrow \infty}\frac{\theta(x)}{x}, \ \ \ \ \
L_2=\limsup_{x\rightarrow \infty}\frac{\theta(x)}{x},
\end{equation*}
and
\begin{equation*}
l_3=\liminf_{x\rightarrow \infty}\frac{\psi(x)}{x}, \ \ \ \ \
L_3=\limsup_{x\rightarrow \infty}\frac{\psi(x)}{x},
\end{equation*}
then $l_1=l_2=l_3$ and $L_1=L_2=L_3$.
\end{theorem}
\begin{proof}
Notice that
\begin{equation*}
\psi(x)=\theta(x)+\theta(x^{1/2})+
\theta(x^{1/3})+...\theta(x^{1/m})\geq \theta(x)
\end{equation*}
where $m\leq log_2x$
\end{proof}
Also,
\begin{equation*}
\psi(x)=\sum_{p\leq x}\left[\frac{\log x}{\log p}\right]\log p\leq
\sum_{p\leq x}\frac{\log x}{\log p} \log p= \log x\pi(x).
\end{equation*}
Thus we have
\begin{equation*}
\theta(x)\leq \psi(x)\leq \pi(x)\log x
\end{equation*}
As a result, we have
\begin{equation*}
\frac{\theta(x)}{x}\leq \frac{\psi(x)}{x}\leq \frac{\pi(x)}{x/\log
x}
\end{equation*}
and we get that $L_2\leq L_3\leq L_1$. We still need to prove that
$L_1 \leq L_2$.
\par Let $\alpha$ be a real number where $0<\alpha<1$, we have
\begin{eqnarray*}
\theta(x)&=&\sum_{p\leq x}\log p\geq \sum_{x^{\alpha}\leq p\leq
x}\log p\\ &>& \sum_{x^{\alpha}\leq p\leq x}\alpha \log x \ \ \
(\log p>\alpha \log x)\\ &=&\alpha log x\{\pi(x)-\pi(x^{\alpha})\}
\end{eqnarray*}
However, $\pi(x^{\alpha})\leq x^{\alpha}$. Hence
\begin{equation*}
\theta(x)>\alpha \log x\{\pi(x)-x^{\alpha}\}
\end{equation*}
As a result,
\begin{equation*}
\frac{\theta(x)}{x} > \frac{\alpha \pi(x)}{x/ \log x}- \alpha
x^{\alpha-1}\log x
\end{equation*}
Since $\lim_{x\rightarrow \infty}\alpha \log x/x^{1-\alpha}=0$, then
\begin{equation*}
L_2\geq \alpha \limsup_{x\rightarrow \infty}\frac{\pi(x)}{x/\log x}
\end{equation*}
As a result, we get that
\begin{equation*}
L_2\geq \alpha L_1
\end{equation*}
As $\alpha \rightarrow 1$, we get $L_2\geq L_1$.
\par Proving that $l_1=l_2=l_3$ is left as an exercise.
We now present an inequality due to Chebyshev about $\pi(x)$.
\begin{theorem}
There exist constants $a1$ and choose $m$ such that $2^{m-1}\leq x\leq 2^m$, we get
that
\begin{equation*}
\theta(x)\leq \theta(2^m)\leq 2^{m+1}\log 2 \leq 4x\log 2
\end{equation*}
and we get $(\ref{1})$ for all $x$.
\par We now prove $(\ref{2})$.
Notice that by Lemma 9, we have that the highest power of a prime
$p$ dividing $N=\frac{(2n)!}{(n!)^2}$ is given by
\begin{equation*}
s_p=\sum_{i=1
1}^{\mu_p}\left\{\left[\frac{2n}{p^i}\right]-2\left[\frac{n}{p^i}\right]\right\}.
\end{equation*}
where $\mu_p=\left[\frac{\log 2n}{\log p}\right]$. Thus we have
$N=\prod_{p\leq 2n}p^{s_p}$. If $x$ is a positive integer then
\begin{equation*}
[2x]-2[x]<2,
\end{equation*}
It means that $[2x]-2[x]$ is $0$ or $1$. Thus $s_p\leq \mu_p$ and
we get
\begin{equation*}
N\leq \prod_{p\leq e2n}p^{\mu_p}.
\end{equation*}
Notice as well that
\begin{equation*}
\psi(2n)=\sum_{p\leq 2n}\left[\frac{\log 2n}{\log p}\right]\log
p=\sum_{p\leq 2n}\mu_p \log p.
\end{equation*}
Hence we get
\begin{equation*}
\log N \leq \psi(2n).
\end{equation*}
Using the fact that $2^{2n}<(2n+1)N$, we can see that
\begin{equation*}
\psi(2n)>2n \log 2-\log (2n+1).
\end{equation*}
Let $x>2$ and put $n=\left[\frac{x}{2}\right]\geq 1$. Thus
$\frac{x}{2}-1__