NT2

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Let $p_n\,$ be the $n t h$ prime. Let $q=2\cdot 3\cdot 5\cdot\cdot\cdot p_n+1$ .

By the proof of NT1, $p_{n+1}\le q$ .

Suppose that $p_n < 2^{2^n}$ for $n = 1,2,..., N$ . It is true for $n = 1$ . The last result gives

$p_{N+1}\le p_1 p_2 ... p_N + 1 < 2^{2+4+...+2^N}+1 < 2^{2^{N+1}}$

which proves it inductively for all $n$ .

Suppose $n\ge 4$ . Let $x$ be such that

$e^{e^{n-1}} < x \le e^{e^n}$ .

Since

$e^{n-1} > 2^n, n>1\,$

it is true that

$e^{e^{n-1}} > 2^{2^n}\,$ .

And so

$\pi(x) \ge \pi(e^{e^{n-1}}) \ge \pi(2^{2^n}) \ge n\,$

Since $\log\log x \le n$ , it is true that

$\pi(x) \ge \log\log x, x\ge 2$

Main Page : Number Theory

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