 Solution

Expand the finite Euler product into geometric series. By unique prime factorization, it is the sum of $1/n$ over positive integers whose prime factors do not exceed $x$, and hence contains every term up to $m=\lfloor x\rfloor$. The harmonic sum satisfiesThe last inequality uses $m+1>x$, including when $x$ is an integer. For each prime number $p$,The telescoping series over all integers $n\ge 2$ bounds the sum of these remainders by $\frac{1}{2}{\sum }_{n\ge 2}[n(n-1){]}^{-1}=1/2$. Taking logarithms of the product bound therefore givesThis explicit constant suffices; no asymptotic theorem about the distribution of prime numbers is needed.