Sum of Ratio of Factorial of Consecutive Primes

srfcp.tex

$$ Michele Venturi \\ dardo82@gmail.com \\ 2019-03-07 $$
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$$ \not\exists \ x \in \mathbb{N} : x \not= 1,\ x \not=P_{n},\ x|P_{n} \\
n \in \mathbb{N}, n \not= 0; P_{n} \in \mathbb{N}, P_{n} \not= 1; p=P_{2n-1},\ q=P_{2n} $$
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$$ \sqrt{2\pi}\ n^{n+\frac12} e^{-n} \le n! \le e\ n^{n+\frac12} e^{-n} \\
\Downarrow \\
\frac{\sqrt{2\pi}\ p^{p+\frac12} e^{-p}}{\sqrt{2\pi}\ q^{q+\frac12} e^{-q}} \le
\frac{p!}{q!} \le \frac{e\ p^{p+\frac12} e^{-p}}{e\ q^{q+\frac12} e^{-q}} \\
\Downarrow \\
\frac{p!}{q!} = \sqrt{\frac{p}{q}} \frac{p^{p}}{q^{q}} e^{q-p} $$
$$ p < q < 2p \Rightarrow \frac12 < \frac{p}{q} < 1 \Rightarrow \frac{\sqrt{2}}{2} < \sqrt{\frac{p}{q}} < 1 $$
$$ p < q < 2p \Rightarrow 0 < q-p < p \Rightarrow 1 < e^{q-p} < e^{p} $$
$$ p < q \Rightarrow p^{p} < q^{q} \Rightarrow \frac{p^{p}}{q^{q}} < 1$$
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$$ \sum_{n=1}^{\infty}\frac{1}{n!} = e $$
$$ \sum_{n=1}^{\infty}\frac{p!}{q!} = \sum_{n=1}^{\infty} \prod_{k=p+1}^{q} k^{-1} \approx e^{-1} $$