# Equivalent statements to PNT.
Most commonly, the prime number theorem states $$
\pi(x) \sim \frac{x}{\log x}
$$
Another form is $$
p_{n} \sim n \log n
$$Here $\pi(x)$ denotes the usual prime counting function, the number of primes less than or equal to $x$; and $p_{n}$ denotes the $n$-th prime. Note that $\pi(p_{n}) = n$.
And recall we write $f(x) \sim g(x)$ if $\lim_{x\to \infty} f(x)/g(x) =1$.
We shall prove the following equivalence, following Apostol,
1. $\pi(x) \sim x / \log x$
2. $\pi(x) \sim x / \log \pi(x)$
3. $p_{n} \sim n \log n$
Suppose $\pi(x) \sim x / \log x$.
Then this means $$
\begin{align*}
& \lim_{x \to \infty} \frac{\pi(x)\log x}{x} = 1 \\
\implies & \lim_{x \to \infty} \log \pi(x) + \log \log x - \log x = 0 \\
\implies & \lim_{x \to \infty} \log x \left( \frac{\log \pi(x)}{\log x} + \frac{\log \log x}{\log x} - 1 \right) = 0 \\
\end{align*}
$$
Since $\log x \to \infty$ as $x \to \infty$, we must have $$
\lim_{x \to \infty} \frac{\log \pi(x)}{\log x} + \frac{\log \log x}{\log x} - 1 = 0
$$
Now, $(\log \log x) / \log x \to 0$ as $x \to \infty$, hence $\log\pi(x) \sim \log x$.
Hence we have $$
\pi(x) \sim \frac{x}{\log x} \sim \frac{x}{\log \pi(x)}.
$$
Now, suppose $\pi(x) \sim x / \log \pi(x)$. Note for each $n$, we have $\pi(p_{n}) = n$. Note as $n \to \infty$ so does $p_{n} \to \infty$.
So, $$
n = \pi(p_{n }) \sim \frac{p_{n}}{\log \pi(p_{n})} = \frac{p_{n}}{\log n}.
$$So we have $$
p_{n} \sim n \log n .
$$
Now, suppose $p_{n} \sim n \log n$, For each $x$, write $n$ such that $p_{n} \le x < p_{n+1}$,
Note as $x$ increases, so does $n$. Also note $n = \pi(p_{n}) = \pi(x)$.
Note $$
\frac{p_{n}}{n \log n } \le \frac{x}{\pi(x) \log\pi(x)} < \frac{p_{n+1}}{n \log n} = \frac{p_{n+1}}{(n+1)\log(n+1)} \frac{(n+1)\log(n+1)}{n \log n}
$$Hence, $x \sim \pi(x)\log(\pi(x))$, or in other words,$$
\pi(x) \sim \frac{x}{\log \pi(x)}.
$$
Lastly, if $\pi(x) \sim x / \log \pi(x)$, then $$
\lim_{x\to \infty} \frac{\pi(x) \log \pi(x)}{x} = 1,
$$
so $$
\begin{align*}
& \lim_{x \to \infty} \log \pi(x) + \log \log \pi(x) - \log x = 0 \\
\implies & \lim_{x \to \infty} \log \pi(x) \left(1+ \frac{\log \log \pi(x)}{\log \pi(x)} - \frac{\log x}{\log \pi (x)}\right) = 0
\end{align*}
$$
Since $\log \pi(x) \to \infty$ as $x \to \infty$, we must have $$
\lim_{x \to \infty} 1 + \frac{\log \log \pi(x)}{\log \pi (x)} - \frac{\log x}{\log \pi(x)} = 0.
$$And since $(\log\log\pi(x)) / \log\pi(x) \to 0$ as $x \to \infty$, we have $$
\log x \sim \log \pi(x).
$$Hence $$
\pi(x) \sim \frac{x}{\log \pi(x)} \sim \frac{x}{\log x}.
$$