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Write $\theta (x)={\sum }_{p\le x}\log p$ for the Chebyshev theta function and $\psi (x)={\sum }_{n\le x}\Lambda (n)$ for the Second Chebyshev function, where $\Lambda$ is the Von Mangoldt function. We first establish the Chebyshev estimate. Every prime $n<p\le 2n$ divides the central binomial coefficient $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$, soApply this at successive powers of two and use monotonicity to obtain $\theta (x)=O(x)$. Sincethe terms $r\ge 2$ contribute $O(\sqrt{x}\log x)$ and therefore $\psi (x)=O(x)$. This Chebyshev estimate from central binomial coefficients is all the information about the distribution of primes needed here.

The Von Mangoldt divisor identity $\log n={\sum }_{d\mid n}\Lambda (d)$ gives the factorial proof of Mertens first theorem. With $N=\lfloor x\rfloor$, interchange the finite sums to obtainComparison of ${\sum }_{n\le N}\log n$ with ${\int }_1^N\log tdt$ gives $\log N!=N\log N-N+O(\log N)$. Dividing the preceding identity by $x$, using $N=x+O(1)$ and the Chebyshev estimate, givesThe contribution of higher prime powers is bounded independently of $x$, becauseThus the Mertens first theorem is

For the square-root logarithmic sum over primes, apply partial summation to $A(t)$ with weight $(\log t{)}^{-1/2}$. The lower endpoint is understood as $A(2^{-})=0$, soThe main terms are $\sqrt{\log x}$ and $\sqrt{\log x}-\sqrt{\log 2}$. The bounded error in $A(t)$ produces a bounded integral, since ${\int }_2^{\infty }dt/(t(\log t{)}^{3/2})<\infty$. ThereforeNeither argument uses the Prime number theorem.