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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.

Write $\vartheta (x)={\sum }_{p\le x}\log p$ for the Chebyshev theta function and $\psi (x)={\sum }_{p^j\le x}\log p$ for the Second Chebyshev function. Throughout, $p$ denotes a prime and all logarithms are natural. Each prime $n<p\le 2n$ divides the central binomial coefficient, soApply this at $n=1,2,4,\dots ,2^{r-1}$ and telescope: $\vartheta (2^r)\le (2^{r+1}-2)\log 2$. Bounding a general $x$ by the next power of two proves $\vartheta (x)=O(x)$. Splitting the prime-counting function at $\sqrt{x}$ now givesThis is the required Chebyshev estimate from central binomial coefficients, with no use of a prime asymptotic.

The higher prime powers also satisfyFor an integer $N$, the prime factorization of a factorial givesThe Stirling formula, or integral comparison for $\log t$, says $\log N!=N\log N-N+O(\log N)$. After division by $N$ we obtain ${\sum }_{p^j\le N}(\log p)/p^j=\log N+O(1)$. MoreoverSubtract this bounded contribution and replace $N$ by $\lfloor x\rfloor$. The factorial proof of Mertens first theorem therefore yields

For fixed $\lambda >1$, apply partial summation to $A(t)$ and $f(t)=(\log t{)}^{\lambda -1}$:The main contribution isThe bounded error in $A(t)$ contributes $O((\log x{)}^{\lambda -1})$ at the endpoint and in the integral, since ${\int }_2^x(\log t{)}^{\lambda -2}dt/t=O_{\lambda }((\log x{)}^{\lambda -1})$. Thus the positive logarithmic moments of reciprocal primes satisfy

Write $\vartheta (x)={\sum }_{p\le x}\log p$ and $\psi (x)={\sum }_{n\le x}\Lambda (n)$ for the two Chebyshev functions, and $\pi (x)$ for the prime-counting function. If $m<p\le 2m$, the prime $p$ divides the central binomial coefficient $\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$. ConsequentlySum this inequality over dyadic integers $m=1,2,4,\dots ,2^{k-1}$. It gives $\vartheta (2^k)\le 2^{k+1}\log 2$. Monotonicity, with $2^{k-1}<x\le 2^k$, gives $\vartheta (x)=O(x)$ for real $x\ge 2$. This is the Chebyshev estimate from central binomial coefficients; no information about the distribution of primes beyond unique factorization has been used.

Split the primes at $\sqrt{x}$. For $p>\sqrt{x}$, $\log p>\frac{1}{2}\log x$, soSince $\sqrt{x}=O(x/\log x)$,We also need $\psi (x)=O(x)$. The contributions of higher prime powers satisfybecause only $k\le \log x/\log 2$ contribute, and $\vartheta (x^{1/k})\ll \sqrt{x}$ for $k\ge 2$.

The Von Mangoldt divisor identity is $\log n={\sum }_{d\mid n}\Lambda (d)$, which follows by factoring $n$ into prime powers. Summing it for $n\le M$, with $M$ a positive integer, givesThe bound already proved makes the last error $O(M)$. Integral comparison for ${\sum }_{n\le M}\log n$, or the Stirling formula, gives $\log (M!)=M\log M-M+O(\log M)$. ThereforeThe contribution from higher prime powers is uniformly bounded:Subtract it, and replace $M$ by $\lfloor x\rfloor$, to obtain the Mertens first theorem

Now fix $\delta >0$. Apply partial summation to $A(t)$ and $f(t)=(\log t{)}^{-1-\delta }$. With the lower endpoint interpreted as $2^{-}$, so that the prime $2$ is included, this givesThe boundary term tends to zero, and the integral converges, since $A(t)\ll \log t$ and ${\int }_2^{\infty }dt/(t(\log t{)}^{1+\delta })<\infty$. Thus the series converges for every positive $\delta$. Keeping the main term of $A(t)$ also gives the useful logarithmically weighted reciprocal-prime tailIndeed, subtract the finite partial sum from the limiting integral: the boundary contributes $-(\log x{)}^{-\delta }$ and the main integral contributes $(1+\delta )(\log x{)}^{-\delta }/\delta$.

Use the Chebyshev estimate from central binomial coefficients. DefineThe first is the Chebyshev theta function; the second counts prime powers with the same logarithmic prime weight. For a positive integer $m$, every prime $m<p\le 2m$ divides $\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$, henceSumming over dyadic intervals yields $\vartheta (2^j)\le 2^{j+1}\log 2$. By monotonicity and rounding $x$ upward to a power of two,

For the lower bound, the central binomial coefficient is the largest of the $2m+1$ coefficients whose sum is $4^m$, soThe exponent of a prime in the central coefficient isEach summand is zero or one. Therefore $\log \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le \psi (2m)$. Higher prime powers contribute onlyCombining the lower bound with $2m$ just below $x$ shows $\vartheta (x)\ge cx$ for sufficiently large $x$, with, for example, $c=(\log 2)/2$.

Let $\pi (x)$ denote the number of primes at most $x$. Since every prime weight is at most $\log x$,For the upper bound, separate the primes at $\sqrt{x}$. The small ones number at most $\sqrt{x}$, and every larger one has weight at least $(\log x)/2$, givingSince $\sqrt{x}=o(x/\log x)$, this provesfor all sufficiently large $n$, for instance with $A=(\log 2)/2$ and $B=8\log 2+1$. This elementary Chebyshev estimate does not assume the Prime number theorem.