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The central binomial coefficient satisfies $4^m/(2m+1)\le \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le 4^m$. Prime factors between $m$ and $2m$ give a dyadic upper bound for the Chebyshev theta function. The prime-power exponents in the factorial quotient are sums of zeros and ones, so the central coefficient gives a linear lower bound for the Second Chebyshev function. Higher prime powers contribute only $O(\sqrt{x}\log x)$; hence the prime-counting function has upper and lower bounds of order $x/\log x$.

The two Chebyshev functions measure the total natural logarithm weight of prime numbers or of prime powers greater than one below a real cutoff $x\ge 0$:Here $p$ ranges over prime numbers, $n$ over positive integers, and $k$ over positive integers. The Chebyshev theta function $\vartheta$, also written $\theta$, weights each prime number once. The Second Chebyshev function $\psi$ weights every positive power of each prime number separately. Both vanish for $0\le x<2$; the inclusive cutoff makes them right-continuous functions and nondecreasing functions, constant between their jumps. ^[ref]

At $x=10$, the prime numbers are $2,3,5,7$, while the extra prime powers greater than one are $4,8,9$. Exactly,At each prime number $p$, both Chebyshev functions jump by $\log p$. At a prime power $p^k$ with $k\ge 2$, only the Second Chebyshev function jumps, by $\log p$. At $6$ and $10$ neither jumps. Notice that the gap is already positive at $5$ and $7$: a shared jump does not remove the earlier contribution from $4$.

Grouping the prime powers by their exponent gives, for $x\ge 2$,The floor function cutoff is exact: $p\ge 2$ forces $2^k\le x$ whenever $p^k\le x$. For each exponent, $p^k\le x$ is equivalent to $p\le x^{1/k}$. An upper cutoff smaller than the lower one means an empty sum. ^[ref]

A positive integer $n$ is decisive when every integer $a$ satisfying $1<a<n$ and $\gcd (a,n)=1$ is a prime number. There are only finitely many decisive numbers: every prime $p<\sqrt{n}$ must divide a decisive $n$, so the Chebyshev theta function gives $\log n\ge \vartheta (\sqrt{n})$, which eventually contradicts the linear lower bound for $\vartheta (\sqrt{n})$.

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

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.

Write the Chebyshev theta function asTaking natural logarithms in part a givesand henceBecause $\sqrt{n}\log n/n\to 0$, the right-hand side is at least $n/2$ for every sufficiently large integer $n$.

Now let $x$ be sufficiently large and set $n=\lfloor x/2\rfloor$ using the floor function. Then $2n\le x$ and $n\ge x/3$, so the monotonicity of $\vartheta$ gives

The Second Chebyshev function is the partial sum of the Von Mangoldt function up to a real cutoff $x\ge 0$:For example, $\psi (3)=\log 6$, $\psi (4)=\log 12$, and $\psi (4.5)=\log 12$. At $4=2^2$ the jump has size $\log 2$; between $4$ and $5$ no new prime power greater than one enters the sum. In contrast, the Chebyshev theta function stays at $\vartheta (4)=\log 6$.

There is also an exact connection with the least common multiple. For $N=\lfloor x\rfloor \ge 1$, each prime number $p\le N$ occurs to the highest exponent $k$ for which $p^k\le N$, both in that least common multiple and in the accumulated Von Mangoldt function weights. ThusThis explains the $\log 2520$ value at $x=10$: the least common multiple is $2^3\cdot 3^2\cdot 5\cdot 7$. The Chebyshev theta function instead gives the natural logarithm of the primorial $2\cdot 3\cdot 5\cdot 7=210$. The additional prime powers retain the highest required multiplicities.

Why is this useful for the distribution of prime numbers? The Prime number theorem has the equivalent formswhere $\pi$ is the prime-counting function and $\sim$ denotes asymptotic equivalence. Thus an arithmetic function defined using prime factorization has average weight tending to one over long initial intervals of the positive integers. These are substantive asymptotic equivalences, not consequences of the definitions or the Von Mangoldt divisor identity alone. ^[ref]

The elementary reason the two Chebyshev functions have the same leading behavior is that their difference is small compared with $x$. Every additional prime power has base $p\le \sqrt{x}$, and for each such prime number its total extra weight is at most $\log x$. There are at most $\lfloor \sqrt{x}\rfloor$ such bases, soThis transfers $\psi (x)\sim x$ to $\vartheta (x)\sim x$, or conversely, once either is established. Partial summation then relates the Chebyshev theta function to the prime-counting function. The estimate controls the extra prime powers; proving the leading value itself is the harder Prime number theorem problem.