In previous Eulerian mathematics, we calculated the sum of the reciprocals of all prime numbers and concluded that the sum of the reciprocals of primes diverges. This also serves as a proof that there are infinitely many prime numbers. In this article, we attempt to calculate the product of all primes. Through a simple technique, we obtain an upper bound for the product of primes (we will also calculate a lower bound in the future), thereby obtaining an upper bound formula for \pi(n). More importantly, this estimation is one of the essential foundations for the elementary proof of Bertrand’s Postulate (which states that there is always a prime between n and 2n). Part of the content of this article is referenced from Proofs from THE BOOK and Introduction to Analytic and Probabilistic Number Theory.
The Product of Primes
As I have mentioned before, the magic of number theory is that it always unexpectedly connects different fields of mathematics. Readers will soon see that the proof in this article has an important connection with combinatorics (albeit a simple one). Regarding the product of primes, we have the following conclusion:
The product of all prime numbers not exceeding n is less than 4^{n-1}.
If we use p to denote a prime number, this can be written as \prod_{p\leq n}p < 4^{n-1}. In fact, more advanced techniques can significantly improve this estimate. From the Prime Number Theorem, we know that \pi(n) \sim \frac{n}{\ln n}, which implies \prod_{p\leq n}p \sim e^n. That is to say, for any \varepsilon > 0, there exists N(\varepsilon) > 0 such that for any n > N(\varepsilon), \prod_{p\leq n}p < (e+\varepsilon)^n. However, the method and estimation used in this article are relatively simple, so we can only obtain the upper bound 4^n. But as readers will discover later, this estimation is sufficient for proving Bertrand’s Postulate.
Now we begin the proof. It can be verified that the conclusion holds for n=2. It is an obvious fact that if the estimate holds for an odd number n, it must also hold for the even number n+1. Therefore, we only need to prove it for n=2m+1. We use mathematical induction, assuming that the conclusion holds for n=1, 2, \dots, 2m. We then split the case for n=2m+1 into two parts: \prod_{p\leq 2m+1}p = \prod_{p\leq m+1}p \prod_{m+1 < p\leq 2m+1}p < 4^m \prod_{m+1 < p\leq 2m+1}p The last inequality is obtained from the inductive hypothesis. Now we need to estimate the other part, which is the product of primes between m+1 and 2m+1. We consider the binomial coefficient: C_{m+1}^{2m+1} = \binom{2m+1}{m+1} = \frac{(2m+1)!}{(m+1)! m!} By definition, C_{m+1}^{2m+1} is an integer, and it is divisible by all prime numbers p such that m+1 < p \leq 2m+1. Therefore, it is clear that: \prod_{m+1 < p\leq 2m+1}p \leq C_{m+1}^{2m+1} Estimating C_{m+1}^{2m+1}, we find: C_{m+1}^{2m+1} \leq 2^{2m} = 4^m This is because C_{m+1}^{2m+1} = C_{m}^{2m+1}, and: C_{m+1}^{2m+1} + C_{m}^{2m+1} \leq \sum_{k=0}^{2m+1} C_{k}^{2m+1} = 2^{2m+1} The exponent 4 is optimal for this method; without changing the estimation approach, a better exponent cannot be obtained from this binomial coefficient. To realize this, one only needs to use Stirling’s formula to perform an asymptotic approximation of C_{m+1}^{2m+1}.
With the estimate for C_{m+1}^{2m+1}, we have: \prod_{p\leq 2m+1}p < 4^m \prod_{m+1 < p\leq 2m+1}p \leq 4^{2m} In other words, the conclusion holds for n=2m+1. Thus, by the principle of mathematical induction, the conclusion holds for all positive integers.
The Number of Primes
With this estimate, we can also provide an upper bound for \pi(n). We only need to notice that: t^{\pi(n)-\pi(t)} < \prod_{p\leq n}p < 4^{n-1} where t is any constant. Taking the logarithm of both sides gives: \pi(n) < \frac{(n-1)\ln 4}{\ln t} + \pi(t) \leq \frac{n\ln 4}{\ln t} + t In number theory, to improve estimation results, the common approach is to introduce some unknown parameters, such as t above, and then adjust these parameters to obtain the optimal estimate. For the inequality above, let f(t) = \frac{(n-1)\ln 4}{\ln t} + t and find its minimum value: f'(t) = -\frac{n\ln 4}{(\ln t)^2 t} + 1 Setting the derivative to 0, we get n\ln 4 = (\ln t)^2 t. This is a transcendental equation that cannot be solved explicitly. An approximate solution can be given through iteration: t_{k+1} = \frac{n\ln 4}{(\ln t_k)^2} Taking t_0 = n, one iteration yields t_1 = \frac{n\ln 4}{(\ln n)^2}. To avoid overly complex solutions, we end our iteration here and, for simplicity, omit the factor \ln 4, taking t = n/(\ln n)^2. Substituting this and simplifying, we get: \pi(n) < \left(\ln 4 + \frac{8\ln\ln n}{\ln n}\right)\frac{n}{\ln n} This estimate is much weaker than the Prime Number Theorem, but it is based on the simplest counting results and is quite satisfactory.
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