# Approximate Value of Nth Prime Number

## Theorem

The $n$th prime number is approximately $n \ln n$.

## Proof

This will be demonstrated by showing that:

$\displaystyle \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln n} = 1$

where $p_n$ denotes the $n$th prime number.

By definition of prime-counting function:

$\map \pi {p_n} = n$

The Prime Number Theorem gives:

$\displaystyle \lim_{x \mathop \to \infty} \dfrac {\map \pi x} {x / \ln x} = 1$

Thus:

$\displaystyle \lim_{x \mathop \to \infty} \dfrac n {p_n / \ln p_n} = 1$
 $\displaystyle \lim_{x \mathop \to \infty} \dfrac {\map \pi x} {x / \ln x}$ $=$ $\displaystyle 1$ Prime Number Theorem $\displaystyle \leadsto \ \$ $\displaystyle \lim_{n \mathop \to \infty} \dfrac n {p_n / \ln p_n}$ $=$ $\displaystyle 1$ Definition of Prime-Counting Function $\text {(1)}: \quad$ $\displaystyle \leadsto \ \$ $\displaystyle \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln p_n}$ $=$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle \lim_{n \mathop \to \infty} \paren {\ln p_n - \ln n - \ln \ln p_n}$ $=$ $\displaystyle 0$ Logarithm of both sides $\displaystyle \leadsto \ \$ $\displaystyle \lim_{n \mathop \to \infty} \ln p_n \paren {1 - \frac {\ln n} {\ln p_n} - \frac {\ln \ln p_n} {\ln p_n} }$ $=$ $\displaystyle 0$ multiplying argument through by $\ln p_n$ $\displaystyle \leadsto \ \$ $\displaystyle \lim_{n \mathop \to \infty} \ln p_n \paren {1 - \frac {\ln n} {\ln p_n} }$ $=$ $\displaystyle 0$ as $\dfrac {\ln n} n \to 0$ $\displaystyle \leadsto \ \$ $\displaystyle \lim_{n \mathop \to \infty} \paren {1 - \frac {\ln n} {\ln p_n} }$ $=$ $\displaystyle 0$ as $\ln p_n \ne 0$ $\displaystyle \leadsto \ \$ $\displaystyle \lim_{n \mathop \to \infty} \frac {\ln n} {\ln p_n}$ $=$ $\displaystyle 1$ as $\ln p_n \ne 0$ $\displaystyle \leadsto \ \$ $\displaystyle \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln n}$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln p_n} \frac {\ln p_n} {\ln n}$ from $(1)$ $\displaystyle$ $=$ $\displaystyle 1$

$\blacksquare$