prime harmonic series diverges - Chebyshev’s proof

Theorem. ${\sum }_{p\text{ prime}}\frac{1}{p}$ diverges.

Proof. (Chebyshev, 1880)
Consider the product

$$\prod _{p\le n}{\left(1-\frac{1}{p}\right)}^{-1}$$

Since ${\left(1-\frac{1}{p}\right)}^{-1}=1+\frac{1}{p}+\frac{1}{p^2}+\mathrm{\cdots }$, we have

$$\prod _{p\le n}{\left(1-\frac{1}{p}\right)}^{-1}=\left(1+\frac{1}{2}+\frac{1}{2^2}+\mathrm{\cdots }\right)\left(1+\frac{1}{3}+\frac{1}{3^2}+\mathrm{\cdots }\right)\left(1+\frac{1}{5}+\frac{1}{5^2}+\mathrm{\cdots }\right)\mathrm{\cdots }$$

So for each $m\le n$, if we expand the above product, $\frac{1}{m}$ will be a term. Thus

$$\prod _{p\le n}{\left(1-\frac{1}{p}\right)}^{-1}\ge \sum _{x=1}^n\frac{1}{x}$$

Taking logarithms, we have

$$\sum _{p\le n}-\ln \left(1-\frac{1}{p}\right)\ge \ln \sum _{x=1}^n\frac{1}{x}$$

But $\ln (1-u)=-u-\frac{u^2}{2}-\frac{u^3}{3}-\mathrm{\cdots }$, so

$$-\ln \left(1-\frac{1}{p}\right)=\frac{1}{p}+\frac{1}{2p^2}+\frac{1}{3p^3}+\mathrm{\cdots }\le \frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\mathrm{\cdots }\le \frac{2}{p}$$

Hence

$$\sum _{p\le n}\frac{2}{p}\ge \sum _{p\le n}\left(-\ln \left(1-\frac{1}{p}\right)\right)\ge \ln \sum _{x=1}^n\frac{1}{x}$$

and thus

$$\sum _{p\le n}\frac{1}{p}\ge \frac{1}{2}\ln \sum _{x=1}^n\frac{1}{x}$$

But the latter series diverges, and the result follows.