In the Wikipedia article about Bertrand's postulate, three proofs are mentioned: Chebyshev's proof, Ramanujan's proof and Erdős's proof. In this post, I give a simplified version of Chebyshev's proof.
In his Mémoire Chebyshev analyzes the expression
\begin{equation*}
T(x) - T\left( \frac{x}{2} \right)- T\left( \frac{x}{3} \right)- T\left( \frac{x}{5} \right)+ T\left( \frac{x}{30} \right)
\end{equation*}
where \( T(x) = \sum_{ 1 \le n \le x} \log n\ \). If one only wants to prove Bertrand's postulate, it is sufficient to analyze instead the simpler expression
\begin{equation}\label{20150927.1}
V(x) = T(x) - T\left( \frac{x}{2} \right)-2 T\left( \frac{x}{3} \right)+ T\left( \frac{x}{6} \right)
\end{equation}
If \( \psi(x) = \sum_{n \le x } \Lambda(n) \) 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*}
V(x) = \psi(x) -\psi\left( \frac{x}{3} \right)+ \psi\left( \frac{x}{5} \right)- \psi\left( \frac{x}{6} \right) + \cdots
\end{equation*}
The coefficients of \( \psi\left( \frac{x}{n} \right) \) in this expression have period 6 and are alternating. This leads immediately to the inequalities
\begin{equation}\label{20150927.3}
V(x) \le \psi(x)
\end{equation}
and
\begin{equation}\label{20150927.4}
\psi(x) -\psi\left( \frac{x}{3} \right) \le V(x)
\end{equation}
To bound \( V(x) \) I use that for all real \( x \ge 1 \)
\begin{equation}\label{20150927.5}
x \log x - x - \log x +1 \le T(x) \le x \log x - x + \log x +1
\end{equation}
From \eqref{20150927.1} and \eqref{20150927.5} one immediately deduces that for all real \( x \ge 6 \)
\begin{equation}\label{20150927.6}
V(x) \le B x + 5 \log x - 1- \log 108
\end{equation}
and
\begin{equation}\label{20150927.7}
B x - 5 \log x - 1 + \log 108 \le V(x)
\end{equation}
with \( B = \frac{1}{6}\log 108 \). We thus immediately have from \eqref{20150927.3} and \eqref{20150927.7} that for all \( x \ge 6 \)
\begin{equation*}
B x - 5 \log x - 1 + \log 108 \le \psi(x)
\end{equation*}
I simplify the inequality a little bit to
\begin{equation}\label{20150927.8}
B x - 5 \log x\le \psi(x)
\end{equation}
One can also check that \eqref{20150927.8} is valid for the wider range \( x \ge 2 \).
Combining \eqref{20150927.4} and \eqref{20150927.6} and using \( - 1 - \log 108 < 0 \), one has for all real \( x \ge 6 \) that
\begin{equation*}
\psi(x) - \psi\left(\frac{x}{3} \right)\le B x + 5 \log x
\end{equation*}
On can easily check that his inequality is more widely valid for all real \( x \ge 1 \). To prove the upper bound, take \( m \in \mathbb{N} \) such that \( 1 \le \frac{x}{3^m} < 3 \). Applying \eqref{20150927.8} several times gives
\begin{align*}
\psi\left(\frac{x}{3} \right)- \psi\left(\frac{x}{3^2} \right) &\le B \frac{x}{3} + 5 \log \frac{x}{3}\\
\psi\left(\frac{x}{3^2} \right)- \psi\left(\frac{x}{3^3} \right) &\le B \frac{x}{3^2} + 5 \log \frac{x}{3^2}\\
\cdots\\
\psi\left(\frac{x}{3^m} \right)- \psi\left(\frac{x}{3^{m+1}} \right) &\le B \frac{x}{3^m} + 5 \log \frac{x}{3^m}\\
\end{align*}
Because \( \frac{x}{3^{m+1}} < 1 \), it follows that \( \psi\left(\frac{x}{3^{m+1}} \right) = 0 \). Adding the inequalities thus gives
\begin{equation*}
\psi(x) \le \frac{3}{2} B x + 5 (m+1) \log x
\end{equation*}
The inequality \( 3^m \le x \) gives \( m \le \frac{\log x }{\log3 } \).
Thus
\begin{equation*}
\psi(x) \le \frac{3}{2} B x + 5 \left( \frac{\log x }{\log3 } + 1\right) \log x
\end{equation*}
I have thus shown that for all \( x \ge 1 \)
\begin{equation*}
\psi_1(x ) \le \psi(x) \le \psi_2(x)
\end{equation*}
with
\begin{equation*}
\psi_1(x ) = B x - 5 \log x
\end{equation*}
and
\begin{equation*}
\psi_2(x ) = \frac{3}{2} B x + 5 \left( \frac{\log x }{\log3 } + 1\right) \log x
\end{equation*}
Using further elementary manipulations, one can deduce from this that
\begin{equation*}
\theta(x) - \theta\left(\frac{x}{2} \right) \ge \psi_1(x) - 2 \psi_2\left( x^{1/2} \right) - \psi_2\left( \frac{x}{2} \right)
\end{equation*}
Because the expression on the right hand side is larger than \( 1 \) for \( x \ge 3400\), the follows that for \( x \ge 3400 \) there is prime \( p \) with
\begin{equation*}
\frac{x}{2} < p \le x
\end{equation*}
Smaller values of \( x \) can be quickly checked by hand or with Mathematica.
Other posts about Bertrand's postulate
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