Chebyshev Bounds and Mertens Theorems — Examples

For $x=10$, the primes are $2,3,5,7$, so $\pi (10)=4$ and $\theta (10)=\log 2+\log 3+\log 5+\log 7.$ By [L2], the prime powers at most $10$ are $2,3,4,5,7,8,9$, so $\psi (10)=3\log 2+2\log 3+\log 5+\log 7.$ These are exactly the first row entries in the displayed table.

The same calculation at $x=20$ and $x=30$ gives the remaining table rows. Up to $30$, the extra contribution in $\psi -\theta$ comes exactly from $4,8,9,16,25,27$, which is the prime-power layer described in [L3].

The table illustrates two qualitative points from the A page: $\psi$ and $\theta$ stay close, but $\psi$ jumps at every prime power while $\theta$ jumps only at the primes.

A direct factorization gives $\left(\genfrac{}{}{0px}{}{20}{10}\right)=184756=2^2\cdot 11\cdot 13\cdot 17\cdot 19.$ The primes in $(10,20]$ are precisely $11,13,17,19$, and each appears with valuation $1$.

The bound in [L1] reads $\frac{4^{10}}{21}\le 184756\le 4^{10},$ so this concrete value sits comfortably inside the abstract central-binomial window used in Chebyshev's proof.

This example shows exactly how the factorization of $\left(\genfrac{}{}{0px}{}{2n}{n}\right)$ detects the interval primes: the large primes appear once, while only small primes contribute higher powers.

The following short certificate covers the entire residual range. Each displayed number is prime, and a prime $p$ is a witness for every integer $n$ with $p\mathrm{/}2<n<p$: $\begin{array}{cc}p& \text{integers }n\text{ covered }\\ 3& 2\\ 5& 3\text{–}4\\ 7& 4\text{–}6\\ 13& 7\text{–}12\\ 23& 12\text{–}22\\ 43& 22\text{–}42\\ 83& 42\text{–}82\\ 163& 82\text{–}162\\ 317& 159\text{–}316\\ 631& 316\text{–}630\end{array}$ Consecutive ranges in the second column overlap or meet consecutively, and their union contains every integer from $2$ through $467$. For each covered $n$, the corresponding prime satisfies $n<p<2n$.

Therefore the finite residual range required by [L1] is closed. This check is evidence for the remaining finitely many cases only; it does not replace the asymptotic part of the theorem.

Summing over the primes up to $10$, $30$, and $100$ gives the four numerical columns in the displayed table.

At each of these three cutoffs, the reciprocal sum is numerically closer to $\log \log x$ than the weighted sum is to $\log x$. This small-range comparison is consistent with the nonzero bounded terms allowed by [L1] and [L2], but it is numerical evidence only and does not compare their asymptotic error strengths.

Multiplying the Euler factors over the primes up to each cutoff and evaluating the comparison term $e^{-\gamma }\mathrm{/}\log x$ gives the displayed table.

The agreement improves across this short range, but the table is numerical evidence only. The exact constant $e^{-\gamma }$ on the A page comes from the Gamma-side analytic computation, not from the data itself.

Let $f(x):=\frac{3}{2}\frac{x}{\log x}(x>e)\mathrm{.}$ By [L1] this is well defined, and $1\cdot \frac{x}{\log x}\le f(x)\le 2\cdot \frac{x}{\log x}$ for every $x>e$, so the two-sided Chebyshev-type bounds hold.

But $\frac{f(x)}{x\mathrm{/}\log x}=\frac{3}{2}$ for every $x>e$, so the ratio does not tend to $1$. Therefore the displayed implication is false.

The two positive functions $F_1(x):=\frac{1}{\log x},F_2(x):=\frac{2}{\log x}$ both satisfy $F_j(x)=\mathrm{\Theta }(1\mathrm{/}\log x)$ as $x\to \mathrm{\infty }$.

Their leading constants are different: one is $1$ and the other is $2$. So a mere $\mathrm{\Theta }(1\mathrm{/}\log x)$ estimate leaves the multiplicative constant free. What Mertens' third theorem for primes adds over that weaker statement is exactly the identification of the constant as $e^{-\gamma }$.