The \(S\)-\(E\) Route to the Chebyshev Bounds for the Prime-Counting Function
April 22, 2026
Abstract
The Chebyshev bounds for the prime-counting function, i.e., \(\pi(x) \asymp x/\ln x\), is established in a new way. This new approach follows the outline \[\pi(x) \;\asymp\; \Bigl(\sum_{p \le x} \sqrt{\tfrac{\ln p}{p}}\Bigr)^{\!2} \;\asymp\; \frac{x}{\ln x}.\] Here, the second \(\asymp\) is derived from the classical estimate by Mertens, i.e., \(\sum_{p \le x} \frac{\ln p}{p} = \ln x + O(1)\); while the first \(\asymp\) is proved by considering the difference \[\Bigl(\sum_{p \le x} \sqrt{\tfrac{\ln p}{p}}\Bigr)^{\!2} \;-\; \sum_{p \le x} \frac{\ln p}{p},\] which is shown as having the same order as \(\pi(x)\).
1 Introduction↩︎
The Chebyshev bounds asserts that \(\pi(x) \asymp x/\ln x\), where \(\pi(x)\) denotes the number of primes less than or equal to \(x\). These bounds were first established by Chebyshev [1] in 1852, and they represent the strongest statement about the distribution of primes that can be obtained without complex analysis. The bounds were later sharpened to the asymptotic \(\pi(x) \sim x/\ln x\) (the prime number theorem) by Hadamard [2] and de la Vallée Poussin [3] in 1896. An elementary proof of the prime number theorem was given by Selberg [4] and Erdős [5] in 1949.
Instead of considering Mertens’ classical summation \(\sum_{p \le x} \frac{\ln p}{p}\), we study the square of a closely related sum, i.e.\(\bigl(\sum_{p \le x} \sqrt{\ln p / p}\bigr)^{2}\), and finally establish the Chebyshev’s bounds for \(\pi(x)\) from our study. The logical deduction of the argument is summarized in Figure 1.
For the ease of expression, we need some convention of notation throughout this paper.
2 Notation↩︎
Throughout this paper, the variable \(x\) is supposed to be greater than or equal to \(2\). This is to make \(1/\ln x\) meaningful.
The relation \(f(x) \asymp g(x)\) means \(f\) and \(g\) have the same order of magnitude, i.e., there exist positive constants \(k\) and \(K\) (independent of \(x\)) such that \(k \cdot g(x) \le f(x) \le K \cdot g(x)\). Clearly, if \(f(x) \asymp g(x)\), then \(g(x) \asymp f(x)\), i.e., the relation \(\asymp\) is symmetric.
Throughout this paper, we use the following notations:
\(\pi(x) =\) the number of primes less than or equal to \(x\);
\(p\) represents a prime (according to context);
\(p_n =\) the \(n\)-th prime;
\(M(x) = \sum_{p \le x} \dfrac{\ln p}{p}\) (\(=\) Mertens’ summation);
\(S(x) = \sum_{p \le x} \sqrt{\dfrac{\ln p}{p}}\) (\(=\) the summation we study);
\(E(x) = \bigl(\sum_{p \le x} \sqrt{\ln p / p}\bigr)^{2} - \sum_{p \le x} \frac{\ln p}{p} = S^{2}(x) - M(x)\) (\(=\) the difference we study);
\(h(x) = \sqrt{\dfrac{x}{\ln x}}\) (\(=\) the square root of the asymptotic function \(x/\ln x\)).
Notice that \[h'(x) \;=\; \frac{1}{2} \cdot \frac{1}{\sqrt{x/\ln x}} \cdot \frac{1 \cdot \ln x - x \cdot \frac{1}{x}}{(\ln x)^{2}} \;>\; 0 \qquad (\text{for } x \ge 3),\] i.e., \(h(x) = \sqrt{x/\ln x}\) is increasing, and so \(1/h(x) = \sqrt{\ln x / x}\) is decreasing.
Under this notation, the outline of our study becomes \[\left. \begin{align} & S^{2}(x) \;\asymp\; \dfrac{x}{\ln x} \\ & \Downarrow \\ & \pi(x) \;\asymp\; S^{2}(x) - M(x) \\ & \Downarrow \\ & \pi(x) \;\asymp\; S^{2}(x) \end{align} \right\} \;\;\Longrightarrow\;\; \pi(x) \;\asymp\; \dfrac{x}{\ln x}. \label{eq:outline}\tag{1}\] Now we establish the above diagram step by step. We begin with the first part of our outline in (1 ).
3 The Estimate \(S^{2}(x) \asymp x/\ln x\)↩︎
This conclusion is equivalent to \(S(x) \asymp \sqrt{x/\ln x}\), which is, in turn, equivalent to the following Lemma 1 and Lemma 2.
Lemma 1. \(S(x) = O\bigl(\sqrt{x/\ln x}\bigr)\).
Lemma 2. \(S(x) > k \cdot \sqrt{x/\ln x}\) for some constant \(k\).
To prove Lemma 1, we need the following two famous results.
Lemma 3 (Abel’s identity, see Theorem 4.2 in [6]). For any arithmetical function \(a(n)\) with \(a(1)=0\) and any continuously differentiable real function \(f(x)\), we have \[\sum_{n \le x} a(n) f(n) \;=\; \Bigl(\sum_{n \le x} a(n)\Bigr) f(x) \;-\; \int_{2}^{x} \Bigl(\sum_{n \le t} a(n)\Bigr) f'(t)\, dt.\]
Clearly, this conclusion is also true if \(\sum_{n \le x}\) is replaced by \(\sum_{p \le x}\), since we could define \(a(n)=0\) for all composite number \(n\)’s.
Lemma 4 (Mertens’ estimate, see [7] or Theorem 4.10 in [6]). \[\sum_{p \le x} \frac{\ln p}{p} \;=\; \ln x + O(1).\]
Now we are ready to prove Lemma 1.
Proof of Lemma 1. We have \[\begin{align} S(x) \;&=\; \sum_{p \le x} \sqrt{\frac{\ln p}{p}} \\ \;&=\; \sum_{p \le x} \frac{\ln p}{p} \cdot \sqrt{\frac{p}{\ln p}} \\ \;&=\; \Bigl(\sum_{p \le x} \frac{\ln p}{p}\Bigr) h(x) - \int_{2}^{x} \Bigl(\sum_{p \le t} \frac{\ln p}{p}\Bigr) h'(t)\, dt && (\text{by Lemma~\ref{lem:abel}})\\ \;&=\; M(x) h(x) - \int_{2}^{x} M(t) h'(t)\, dt \\ \;&=\; \bigl(\ln x + O(1)\bigr) h(x) - \int_{2}^{x} \bigl(\ln t + O(1)\bigr) h'(t)\, dt && (\text{by Lemma~\ref{lem:mertens}})\\ \;&=\; h(x) \ln x + O\bigl(h(x)\bigr) - \int_{2}^{x} \ln t \cdot h'(t)\, dt + O\!\left(\int_{2}^{x} h'(t)\, dt\right) \\ \;&=\; \sqrt{\frac{x}{\ln x}} \cdot \ln x - \int_{2}^{x} \ln t \cdot h'(t)\, dt + O\bigl(h(x)\bigr) + O\bigl(h(x)-h(2)\bigr) \\ \;&=\; \sqrt{x \cdot \ln x} - \int_{2}^{x} \ln t \cdot h'(t)\, dt + O\bigl(h(x)\bigr). \end{align}\] We rewrite the above as \[S(x) \;=\; \sqrt{x \cdot \ln x} \;-\; \int_{2}^{x} \ln t \cdot h'(t)\, dt \;+\; O\bigl(h(x)\bigr). \label{eq:Sx}\tag{2}\]
Now we estimate the integral in the middle: \[\begin{align} \int_{2}^{x} \ln t \cdot h'(t)\, dt \;&=\; -\int_{2}^{x} \frac{h(t)}{t}\, dt + \ln t \cdot h(t)\Big|_{2}^{x} && (\text{integration by parts}) \\ \;&=\; -\int_{2}^{x} \frac{\sqrt{t / \ln t}}{t}\, dt + \ln t \cdot \sqrt{\frac{t}{\ln t}}\,\Big|_{2}^{x} \\ \;&=\; -\int_{2}^{x} \frac{1}{\sqrt{t \cdot \ln t}}\, dt + \sqrt{t \cdot \ln t}\,\Big|_{2}^{x} \\ \;&=\; -\int_{2}^{x} \frac{1}{\sqrt{t \cdot \ln t}}\, dt + \sqrt{x \cdot \ln x} \;+\; \text{constant} \\ \;&=\; O\!\left(\sqrt{\frac{x}{\ln x}}\right) + \sqrt{x \cdot \ln x} + \text{constant}. \end{align}\] That is, \[\int_{2}^{x} \ln t \cdot h'(t)\, dt \;=\; O\!\left(\sqrt{x/\ln x}\right) + \sqrt{x \cdot \ln x} + \text{constant}. \label{eq:integral}\tag{3}\] The last equality is derived from the following limit, computed by L’Hospital’s rule: \[\lim_{x \to \infty} \frac{\int_{2}^{x} \frac{1}{\sqrt{t \cdot \ln t}}\, dt}{\sqrt{x / \ln x}} \;=\; \lim_{x \to \infty} \frac{\dfrac{1}{\sqrt{x \cdot \ln x}}}{\dfrac{1}{2} \cdot \dfrac{1}{\sqrt{x/\ln x}} \cdot \dfrac{\ln x - 1}{(\ln x)^{2}}} \;=\; \lim_{x \to \infty} \frac{2}{1 - \frac{1}{\ln x}} \;=\; 2.\]
Lastly, plugging Equation (3 ) into Equation (2 ), we get \[\begin{align} S(x) \;&=\; \sqrt{x \cdot \ln x} - \int_{2}^{x} \ln t \cdot h'(t)\, dt + O\bigl(h(x)\bigr) \\ \;&=\; \sqrt{x \cdot \ln x} - \Bigl(O\bigl(\sqrt{x/\ln x}\bigr) + \sqrt{x \cdot \ln x} + \text{constant}\Bigr) + O\!\left(\sqrt{\frac{x}{\ln x}}\right) \\ \;&=\; O\!\left(\sqrt{x/\ln x}\right). \end{align}\] This completes the proof of Lemma 1. ◻
Now let us focus on Lemma 2.
Proof of Lemma 2. For any fixed constant \(c\), consider the partial sum \[S(x) - S(x/c) \;=\; \sum_{x/c < p \le x} \sqrt{\frac{\ln p}{p}}.\] Since \(x/c < p\), we have \[\sqrt{\frac{\ln p}{p}} \;\le\; \sqrt{\frac{\ln(x/c)}{x/c}} \qquad \bigl(\sqrt{\ln x / x} = 1/h(x) \text{ is decreasing}\bigr),\] so \[\frac{\ln p}{p} \;\le\; \sqrt{\frac{\ln(x/c)}{x/c}} \cdot \sqrt{\frac{\ln p}{p}},\] thus \[\sum_{x/c < p \le x} \frac{\ln p}{p} \;\le\; \sqrt{\frac{\ln(x/c)}{x/c}} \cdot \Bigl(\sum_{x/c < p \le x} \sqrt{\frac{\ln p}{p}}\Bigr),\] i.e., \[\frac{\sum_{x/c < p \le x} \frac{\ln p}{p}}{\sqrt{\ln(x/c)/(x/c)}} \;\le\; \sum_{x/c < p \le x} \sqrt{\frac{\ln p}{p}} \;=\; S(x) - S(x/c) \;<\; S(x).\] We can choose a suitable \(c\) such that \(\sum_{x/c < p \le x} \frac{\ln p}{p} > 1\) (see (4 ) at the end of this proof). Then we have \[S(x) \;>\; \frac{\sum_{x/c < p \le x} \frac{\ln p}{p}}{\sqrt{\ln(x/c)/(x/c)}} \;>\; \frac{1}{\sqrt{\ln(x/c)/(x/c)}} \;=\; \sqrt{\frac{x/c}{\ln(x/c)}} \;=\; \frac{1}{\sqrt{c}} \cdot \sqrt{\frac{x}{\ln x - \ln c}} \;>\; \frac{1}{\sqrt{c}} \cdot \sqrt{\frac{x}{\ln x}}.\] This is the conclusion of Lemma 2.
Finally, we show that a constant \(c\) could be chosen such that \(\sum_{x/c < p \le x} \frac{\ln p}{p} > 1\). In fact, \[\sum_{x/c < p \le x} \frac{\ln p}{p} \;=\; M(x) - M(x/c) \;=\; \ln x - \ln(x/c) + O(1) \;=\; \ln c + O(1), \label{eq:gap}\tag{4}\] where the second equality is by Lemma 4. Clearly, we can choose \(c\) such that \[\sum_{x/c < p \le x} \frac{\ln p}{p} \;=\; \ln c + O(1) \;>\; 1\] (since this \(O(1)\) does not depend on \(c\)). The proof of Lemma 2 is completed. ◻
Now that the first part of the outline in (1 ), i.e., \(S^{2}(x) \asymp x/\ln x\), is established, we can proceed to its second part.
4 The Estimate \(\pi(x) \asymp S^{2}(x) - M(x)\)↩︎
Lemma 5. \(\pi(x) \asymp S^{2}(x) - M(x)\).
Proof of Lemma 5. The main idea is to study the difference function \[E(x) \;=\; S^{2}(x) - M(x).\] Recall that \(p_n\) denotes the \(n\)-th prime. And, for convenience, we denote the evaluations \(S(p_n)\), \(M(p_n)\), and \(E(p_n)\) by \(S_n\), \(M_n\), and \(E_n\), respectively.
Then \[\sum_{p \le p_n} \sqrt{\frac{\ln p}{p}} \;=\; \sum_{p \le p_{n-1}} \sqrt{\frac{\ln p}{p}} + \sqrt{\frac{\ln p_n}{p_n}},\] i.e., \(S(p_n) = S(p_{n-1}) + a_n\), where \(a_n = \sqrt{\ln p_n / p_n}\), i.e., \(S_n = S_{n-1} + a_n\). Similarly, \(M_n = M_{n-1} + a_n^{2}\). Therefore \[\begin{align} S_n^{2} - M_n \;&=\; (S_{n-1} + a_n)^{2} - (M_{n-1} + a_n^{2}) \\ \;&=\; S_{n-1}^{2} + 2 a_n S_{n-1} + a_n^{2} - M_{n-1} - a_n^{2} \\ \;&=\; S_{n-1}^{2} - M_{n-1} + 2 a_n S_{n-1}, \end{align}\] i.e., \(E_n = E_{n-1} + 2 a_n S_{n-1}\). Hence \[\begin{align} E_n \;&=\; E_{n-1} + 2 a_n S_{n-1}, \\ E_{n-1} \;&=\; E_{n-2} + 2 a_{n-1} S_{n-2}, \\ \;&\;\;\vdots \\ E_2 \;&=\; E_1 + 2 a_2 S_1. \end{align}\] Summing up all the above equalities, we get \[E_n \;=\; E_1 + 2 a_n S_{n-1} + 2 a_{n-1} S_{n-2} + \cdots + 2 a_2 S_1. \label{eq:En}\tag{5}\]
Now we estimate \(2 a_n S_{n-1}\). Recall that, by Lemma 1, we have \(S(x) \asymp \sqrt{x/\ln x}\). So we have \[\frac{a_n}{S_n} \;\asymp\; \frac{\sqrt{\ln p_n / p_n}}{\sqrt{p_n / \ln p_n}} \;=\; \frac{\ln p_n}{p_n} \;\longrightarrow\; 0 \qquad (\text{as } n \to \infty),\] thus \(a_n / S_n \ll 1\), so \(a_n \ll S_n\), and, consequently, \(S_{n-1} = S_n - a_n \asymp S_n\).
On the other hand, since \(S(x) \asymp \sqrt{x/\ln x}\), we have \(S_n \asymp \sqrt{p_n / \ln p_n} = 1/a_n\), thus \(a_n S_n \asymp 1\). Combining the above observation, i.e., \(S_{n-1} \asymp S_n\), we conclude that \[a_n S_{n-1} \;\asymp\; a_n S_n \;\asymp\; 1,\] i.e., there exist positive constants \(k\) and \(K\) such that \(k \le a_n S_{n-1} \le K\) (holds for any \(n\) sufficiently large).
Putting this result into Equation (5 ), we conclude that \(E_n \asymp n\), i.e., \[S_n^{2} - M_n \;\asymp\; n,\] i.e., \(S^{2}(p_n) - M(p_n) \asymp n = \pi(p_n)\). Hence \[S^{2}(x) - M(x) \;\asymp\; \pi(x).\] The proof of Lemma 5 is completed. ◻
5 Summary and the Chebyshev Bounds↩︎
In the previous section, we see that \[\pi(x) \;\asymp\; S^{2}(x) - M(x).\] This implies \(\pi(x) \asymp S^{2}(x)\). (Notice that \(M(x) = \ln x + O(1) = o(x/\ln x) = o(S^{2}(x))\), where the last equality is by \(S^{2}(x) \asymp x/\ln x\).)
Combining \(\pi(x) \asymp S^{2}(x)\) and \(S^{2}(x) \asymp x/\ln x\) (Lemma 1 together with Lemma 2), we finally have the following theorem.
Theorem 1 (Chebyshev bounds). \[\pi(x) \;\asymp\; \dfrac{x}{\ln x}.\]
Acknowledgments↩︎
The author thanks Dr.Buliao Wang for his sustained mentorship in analytic number theory and for checking an early version of this proof.