Mills' Theorem/Lemma 2

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Lemma for Mills' Theorem

Let:

$\N$ denote the set of all natural numbers

$\Bbb P$ denote the set of all prime numbers.

Let $p_n$ be the $n$th prime number.

From Difference between Consecutive Primes:

$p_{n + 1} - p_n < K {p_n}^{5 / 8}$

where $K$ is an unknown but fixed positive integer.

Let $P_0 > K^8$ be a prime number.

By Lemma 1, there exists an infinite sequence of primes:

$P_0, P_1, P_2, \ldots$

such that:

$\forall n \in \N_{>0}: {P_n}^3 < P_{n + 1} < \paren {P_n + 1}^3 - 1$

Let us define a mapping $u: \N \to \Bbb P$ as:

$\forall n \in \N: \map u n = {P_n}^{3^{-n} }$

Then we have that:

$\forall n \in \N_{>0}: \map u {n + 1} > \map u n$

Proof

\(\ds \map u {n + 1}\)

\(=\)

\(\ds {P_{n + 1} }^{3^{-\paren {n + 1} } }\)

\(\ds \)

\(>\)

\(\ds \paren { {P_n}^3}^{3^{-n - 1} }\)

because $P_{n + 1} > {P_n}^3$

\(\ds \)

\(=\)

\(\ds {P_n}^{3 \times 3^{-n-1} }\)

\(\ds \)

\(=\)

\(\ds {P_n}^{3^{-n} }\)

\(\ds \)

\(=\)

\(\ds \map u n\)

$\blacksquare$