Mertens first theorem

The prime sum ${\sum }_{p\le y}(\log p)/p=\log y+O(1)$ is the first Mertens theorem. It controls the mean logarithmic size in a truncated Euler-product lower bound.

Put $A(t)={\sum }_{p\le t}(\log p)/p$. The Mertens first theorem gives $A(t)=\log t+O(1)$. Partial summation yieldsThe main terms sum to $2\sqrt{\log x}$ up to a constant. The error integral is bounded because ${\int }_2^{\infty }dt/(t(\log t{)}^{3/2})$ converges. Thus a prime sum with a fractional logarithmic weight follows from an elementary bounded-error estimate, without the Prime number theorem.

For fixed $\lambda >1$, integrate $(\log t{)}^{\lambda -1}$ against $A(t)={\sum }_{p\le t}(\log p)/p$. The Mertens first theorem gives $A(t)=\log t+O(1)$. Partial summation makes the main integral $(\log x{)}^{\lambda }/\lambda$; the bounded error in $A$ contributes at most a constant times $(\log x{)}^{\lambda -1}$. The restriction $\lambda >1$ ensures that integrating the derivative $(\lambda -1)(\log t{)}^{\lambda -2}/t$ has this bound.

The prime factorization of $N!$ gives $\log N!={\sum }_{p^j\le N}\lfloor N/p^j\rfloor \log p$. The Chebyshev estimate bounds the error on replacing each floor by its argument by $O(N)$. Dividing by $N$ and using the Stirling formula gives ${\sum }_{p^j\le N}(\log p)/p^j=\log N+O(1)$. The contribution of $j\ge 2$ is bounded by the convergent series ${\sum }_{n\ge 2}(\log n)/(n(n-1))$. This proves the Mertens first theorem without the Prime number theorem.

For fixed $\delta >0$, the Mertens first theorem and partial summation imply convergence of ${\sum }_{p}1/(p(\log p{)}^{\delta })$ and, more precisely, the displayed tail asymptotic with error $O((\log x{)}^{-1-\delta })$. Write $A(t)={\sum }_{p\le t}(\log p)/p=\log t+O(1)$ and integrate against $(\log t{)}^{-1-\delta }$. The boundary at infinity vanishes; combining the finite boundary with the integral gives the coefficient $1/\delta$.