Theorem. $\sum_{p{ prime}}\frac{1}{p}$ diverges.

Proof. (Chebyshev, 1880)
Consider the product $$ \prod_{p\leq n} \left(1-\frac{1}{p}\right)^{-1 $$ Since $\left(1-\frac{1}{p}\right)^{-1}=1+\frac{1}{p}+\frac{1}{p^2}+\cdots$ , we have $$ \prod_{p\leq n} \left(1-\frac{1}{p}\right)^{-1}=\left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots\right)\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots\right)\left(1+\frac{1}{5}+\frac{1}{5^2}+\cdots\right)\cdot $$ So for each $m\leq n$ , if we expand the above product, $\frac{1}{m}$ will be a term. Thus $$ \prod_{p\leq n}\left(1-\frac{1}{p}\right)^{-1}\geq\sum_{x=1}^n \frac{1}{x $$ Taking logarithms, we have $$ \sum_{p\leq n}-\ln\left(1-\frac{1}{p}\right)\geq\ln\sum_{x=1}^n \frac{1}{x $$ But $\ln(1-u)=-u-\frac{u^2}{2}-\frac{u^3}{3}-\cdots$ , so $$ -\ln\left(1-\frac{1}{p}\right)=\frac{1}{p}+\frac{1}{2p^2}+\frac{1}{3p^3}+\cdots\leq \frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\cdots\leq \frac{2}{p $$ Hence $$ \sum_{p\leq n} \frac{2}{p}\geq \sum_{p\leq n}\left(-\ln\left(1-\frac{1}{p}\right)\right)\geq \ln\sum_{x=1}^{n}\frac{1}{x $$ and thus $$ \sum_{p\leq n}\frac{1}{p}\geq \frac{1}{2}\sum_{x=1}^{n}\frac{1}{x $$ But the latter series diverges, and the result follows.