A good motivation why Chebyshev uses \eqref{HNTPeq:1} can be found in this master thesis. Namely, if one defines \( \psi(x) = \sum_{n \le x } \Lambda(n) \) with \( \Lambda \) the von Mangoldt function, then \begin{equation*} T(x) = \sum_{n=1}^{\infty} \psi\left(\frac{x}{n} \right) \end{equation*} and thus \begin{equation*} \psi(x) = \sum_{n=1}^{\infty} \mu(n)\ T\left(\frac{x}{n} \right) \end{equation*} See Murty's book for example for these formulas. Because \( T(x) \approx x \log x - x \), \(\psi(x) \) can be approximated as \begin{equation*} \psi(x) \approx \sum_{n\le x} \mu(n)\left( \frac{x}{n} \log\frac{x}{n} - \frac{x}{n}\right) \approx \sum_{n\le x} \frac{\mu(n)}{n} \left(x \log x - x \right) - \left( \sum_{n\le x} \frac{\mu(n)}{n} \log n \right)x \end{equation*} The prime number theorem, which was not proved yet at the time of Chebyshev, states that \( \psi(x) \sim x \). It thus seems natural to use some kind of approximation to the Möbius function \( \tilde\mu \approx \mu\) so that \( \tilde\mu(n) = 0 \) for large \(n\) and \( \sum \frac{\tilde\mu(n)}{n} = 0 \) in order to remove the term \( x \log x \). Chebyshev took \( \tilde\mu(1) = 1,\ \tilde\mu(2) = -1,\ \tilde\mu(3) = -1,\ \tilde\mu(5) = -1,\ \tilde\mu(30) = 1,\) and other \( \tilde\mu(n) = 0\). For this choice, it follows indeed that \begin{equation*} 1 - \frac{1}{2} - \frac{1}{3}- \frac{1}{5}+\frac{1}{30} = 0 \end{equation*} To get close to the prime number theorem, one also would like that \( - \sum\frac{\tilde\mu(n)}{n} \log n \approx 1 \). Chebyshev's choice gives \( - \sum\frac{\tilde\mu(n)}{n} \log n \approx 0.92 \) which is quite close to 1. In the following, I will take instead \( \tilde\mu(1) = 1,\ \tilde\mu(2) = -2\), and other \( \tilde\mu(n) = 0\). For this simpler choice, \begin{equation*} \sum\frac{\tilde\mu(n)}{n} = 1 - \frac{2}{2} = 0 \end{equation*} and \begin{equation*} \sum\frac{\tilde\mu(n)}{n} \log n \approx 0.69 \end{equation*} Because 0.69 is farther from 1 than Chebyshev's value 0.92, one thus expects this simpler choice for \( \tilde\mu\) to give weaker results than Chebyshev's results. I now follow Chebyshev's reasoning from his paper but on this simpler choice of \( \tilde \mu \). Additionally, I use Landau's large \( O \) notation whereas Chebyshev's uses explicit inequalities. I thus study the function \begin{equation*} T(x) - 2\ T\left( \frac{x}{2} \right) \end{equation*} Because \(T(x) = \sum_{n=1}^{\infty} \psi\left(\frac{x}{n} \right) \), one has \begin{equation*} T(x) - 2\ T\left( \frac{x}{2} \right) = \psi(x) - \psi\left(\frac{x}{2} \right)+\psi\left(\frac{x}{3} \right)-\psi\left(\frac{x}{4} \right)+\cdots \end{equation*} This series is alternating and decreasing, thus \begin{equation}\label{eq:2} \psi(x) - \psi\left(\frac{x}{2} \right)\le T(x) - 2\ T\left( \frac{x}{2} \right) \le \psi(x) \end{equation} Because \( T(x) = x \log x - x + O(\log x ) \), it follows that \begin{align*} T(x) - 2\ T\left( \frac{x}{2} \right) &= x \log x - x + O(\log x ) -2 \left( \frac{x}{2} \log\frac{x}{2} - \frac{x}{2} + O\left( \log \frac{x}{2} \right) \right)\\ & = - x + x \cdot \log 2 + x + O(\log x ) \\ &= \log 2 \cdot x + O(\log x ) \end{align*} The terms involving \( x \log x \) have cancelled. This is not a coincidence, but a consequence of the choice of \( \tilde \mu \). The function \( \psi(x) \) has thus the following lower bound \begin{equation}\label{eq:3} \log 2 \cdot x + O(\log x ) \le \psi(x) \end{equation} To obtain an upper bound , one can proceed as follows (Chebyshev uses a different argument) \begin{align*} \psi(x) - \psi\left(\frac{x}{2} \right) & \le \log 2 \cdot x + O(\log x )\\ \psi\left(\frac{x}{2} \right)- \psi\left(\frac{x}{4} \right) & \le \log 2 \cdot \frac{x}{2} + O(\log x )\\ \psi\left(\frac{x}{4} \right)- \psi\left(\frac{x}{8} \right) & \le \log 2 \cdot \frac{x}{4} + O(\log x )\\ \cdots \end{align*} Because \( \psi\left(\frac{x}{2^m} \right) = 0 \) if \( m > \frac{\log x}{\log 2} \), adding the above inequalities gives the upper bound \begin{equation}\label{eq:4} \psi(x) \le 2 \cdot \log 2 \cdot x + O(\log^2 x )\\ \end{equation}

__Comments__

- From \eqref{eq:3} and \eqref{eq:4} it follows that \begin{equation*} \log 2 \le \liminf_{x \to \infty} \frac{\psi(x)}{x} \le\limsup_{x \to \infty} \frac{\psi(x)}{x}\le 2 \log 2 \end{equation*} This is weaker than the prime number theorem \begin{equation*} \lim_{x \to \infty} \frac{\psi(x)}{x} =1 \end{equation*}
- The inequalities \eqref{eq:3} and \eqref{eq:4} are not strong enough to prove Bertrand's postulate. Indeed, I can only deduce that \begin{align*} \psi(x) - \psi\left(\frac{x}{2} \right) &\ge \log 2 \cdot x - 2 \log 2 \cdot \frac{x}{2} + O(\log^2 x )\\ &\ge O(\log^2 x ) \end{align*} because the coefficient of the factor x has cancelled. Ramanujan has modified the reasoning above so that Bertrand's postulate does follow, see Murty p 38.
- In the next post, I will repeat the argument using explicit inequalities instead of using the big \( O \) notation.

__Other posts about Bertrand's postulate__

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