# prime harmonic series diverges - Chebyshev’s proof

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

 $\prod_{p\leq n}\left(1-\frac{1}{p}\right)^{-1}=\left(1+\frac{1}{2}+\frac{1}{2^% {2}}+\cdots\right)\left(1+\frac{1}{3}+\frac{1}{3^{2}}+\cdots\right)\left(1+% \frac{1}{5}+\frac{1}{5^{2}}+\cdots\right)\cdots$

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

 $\sum_{p\leq n}\frac{1}{p}\geq\frac{1}{2}\ln\sum_{x=1}^{n}\frac{1}{x}$

But the latter series diverges, and the result follows.

Title prime harmonic series diverges - Chebyshev’s proof PrimeHarmonicSeriesDivergesChebyshevsProof 2013-03-22 16:23:48 2013-03-22 16:23:48 rm50 (10146) rm50 (10146) 9 rm50 (10146) Theorem msc 11A41