Approximate Value of Nth Prime Number

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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 $(1)$
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$


Sources