# 素数的倒数之和

## 证明一

${\displaystyle \ln \left(\sum _{n=1}^{\infty }{\frac {1}{n}}\right)=\ln \left(\prod _{p}{\frac {1}{1-p^{-1}}}\right)=\sum _{p}\ln \left({\frac {1}{1-p^{-1}}}\right)=\sum _{p}-\ln(1-p^{-1})}$
${\displaystyle =\sum _{p}\left({\frac {1}{p}}+{\frac {1}{2p^{2}}}+{\frac {1}{3p^{3}}}+\cdots \right)=\left(\sum _{p}{\frac {1}{p}}\right)+\sum _{p}{\frac {1}{p^{2}}}\left({\frac {1}{2}}+{\frac {1}{3p}}+{\frac {1}{4p^{2}}}+\cdots \right)}$
${\displaystyle <\left(\sum _{p}{\frac {1}{p}}\right)+\sum _{p}{\frac {1}{p^{2}}}\left(1+{\frac {1}{p}}+{\frac {1}{p^{2}}}+\cdots \right)=\left(\sum _{p}{\frac {1}{p}}\right)+\left(\sum _{p}{\frac {1}{p(p-1)}}\right)}$
${\displaystyle =\left(\sum _{p}{\frac {1}{p}}\right)+C}$

${\displaystyle {\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{11}}+\cdots =\ln \ln(+\infty ).}$

## 证明二

${\displaystyle \sum _{k=1}^{\infty }{1 \over p_{k}}=c.}$

${\displaystyle \sum _{k=1}^{\infty }{1 \over p_{i+k}}<{1 \over 2}.}$

${\displaystyle N(x)\leq 2^{i}{\sqrt {x}}\,}$

${\displaystyle x-N(x)<\sum _{k=1}^{\infty }{x \over p_{i+k}}<{x \over 2},}$

${\displaystyle {x \over 2}