In the previous article, I briefly discussed the “Riemann \zeta function” and the “Golden Key” in number theory through the lens of Euler’s mathematics. In fact, the connection between this “Golden Key” and many problems has already been established. In other words, the “Golden Key” has been inserted into the corresponding “keyhole,” and the mathematician’s job is to “turn” this key, thereby opening the door to mathematics!
Next, let us see how to prove that the sum of the reciprocals of all prime numbers diverges. Before getting to the main topic, we need a lemma:
If every term of an infinite sequence \{a_n\} is greater than 0, then the convergence or divergence of \sum\limits_{n=1}^{\infty} a_n is the same as that of \prod\limits_{n=1}^{\infty} \left(1+a_n\right). In other words, they are necessary and sufficient conditions for each other!
How do we prove this? Let S=a_1+a_2+\dots+a_n and T=\left(1+a_1\right)\left(1+a_2\right)\dots\left(1+a_n\right). First, we prove T > 1+S. This should be quite simple, because \left(1+a_1\right)\left(1+a_2\right)=1+a_1+a_2+a_1 a_2 > 1+\left(a_1+a_2\right). Therefore, \left(1+a_1\right)\left(1+a_2\right)\left(1+a_3\right) > [1+\left(a_1+a_2\right)]\left(1+a_3\right) > 1+\left(a_1+a_2+a_3\right). This can be continued by induction. Next, let us recall an inequality of means: if x_i are all positive numbers, then: \frac{x_1+x_2+\dots+x_n}{n} \geq \sqrt[n]{x_1 x_2 \dots x_n} Or written as: x_1 x_2 \dots x_n \leq \left(\frac{x_1+x_2+\dots+x_n}{n}\right)^n In this case: \left(1+a_1\right)\left(1+a_2\right)\dots\left(1+a_n\right) \leq \left(\frac{n+a_1+a_2+\dots+a_n}{n}\right)^n =\left(1+\frac{S}{n}\right)^n The last step, from the definition of the base of the natural logarithm e, gives \left(1+\frac{S}{n}\right)^n < e^S. Thus, we have actually proven: T < e^S. Combining these, we get: 1+S < T < e^S
Therefore, the range of T is entirely determined by S. Obviously, if S is finite, T cannot tend to infinity! Extending n to infinity clearly holds as well. Q.E.D. As a side note: it can be verified that e^S is a very good approximation of T, with an accuracy far exceeding (1+S). Thus, in many cases, one can directly use T \approx e^S.
Now, let us bring out our “Golden Key” — \zeta \left(s\right)=\prod\limits_{p} \left(1-p^{-s}\right)^{-1}. Here we only use the case s=1, and we truncate the primes on the right side up to p, namely: \begin{aligned} &\left(1+\frac{1}{2}+\frac{1}{2^2}+\dots\right)\left(1+\frac{1}{3}+\frac{1}{3^2}+\dots\right)\dots\left(1+\frac{1}{p}+\frac{1}{p^2}+\dots\right)\\ =&\frac{2}{2-1}\cdot \frac{3}{3-1}\cdot \frac{5}{5-1}\cdot \dots\cdot \frac{p}{p-1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{p}+\dots\\ > &1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{p} \\ > &\ln\left(p+1\right) \end{aligned}
Let p_n denote the n-th prime number. It is not difficult to prove that: \frac{p_n}{p_n -1} < 1+\frac{1}{p_{n-1}}.
Thus: \begin{aligned} \ln\left(p+1\right) < &\frac{2}{2-1}\cdot \frac{3}{3-1}\cdot \frac{5}{5-1}\cdot \dots\cdot \frac{p}{p-1} \\ < &2\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right)\dots\left(1+\frac{1}{p}\right)\\ < &2e^{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{p}\right)} \end{aligned}
Let Q=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{p}. We then have \ln\left(p+1\right) < 2e^Q, which implies: Q > \ln \ln \left(p+1\right) -\ln2
Since there are infinitely many primes (the number of primes is infinite and they are increasing), the sum of the reciprocals of all primes diverges! Q.E.D.
Finally, based on the earlier point that “e^S is a very good approximation of T,” we can consider \ln \ln p to be a very good approximation of Q.
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