[Euler's Mathematics] The Sum of Reciprocals of Primes

[欧拉数学]素数倒数之和 苏剑林 · 2011-11-19 · Mathematics prime number euler mathematics

In the previous post, I used Euler's methods to give a simple account of the "Riemann zeta function" and the "golden key" in number theory. In fact, the links between this "golden key" and many problems have already been established—in other words, the "golden key" has already been inserted into the corresponding "keyhole," and the mathematician's job is to "turn" this key and thereby open the door to mathematics!

Next let's see how to prove that the sum of the reciprocals of all primes diverges. Before getting to the main topic, we need a lemma:

If every term of the infinite sequence ${a_n}$ is greater than 0, then $\sum\limits_{n=1}^{\infty} a_n$ and $\prod\limits_{n=1}^{\infty} \left(1+a_n\right)$ have the same convergence behavior. In other words, each is a necessary and sufficient condition for the other!

How do we prove this? Write $S=a_1+a_2+...+a_n,T=\left(1+a_1\right)\left(1+a_2\right)...\left(1+a_n\right)$. First we prove T > 1+S, which should be easy, since $\left(1+a_1\right)\left(1+a_2\right)=1+a_1+a_2+a_1 a_2 > 1+\left(a_1+a_2\right)$, so $\left(1+a_1\right)\left(1+a_2\right)\left(1+a_3\right) > [1+\left(a_1+a_2\right)]\left(1+a_3\right) > 1+\left(a_1+a_2+a_3\right)$, and we just keep going by induction. Next let's recall an inequality about averages: if $x_i$ are all positive numbers, then:

$$\frac{x_1+x_2+...+x_n}{n} \geq \sqrt[n]{x_1 x_2...x_n}$$

or equivalently

$$x_1 x_2...x_n \leq \left(\frac{x_1+x_2+...+x_n}{n}\right)^n$$

In that case

$$\left(1+a_1\right)\left(1+a_2\right)...\left(1+a_n\right) \leq \left(\frac{n+a_1+a_2+...+a_n}{n}\right)^n =\left(1+\frac{S}{n}\right)^n$$

The last step follows from the definition of e, the base of the natural logarithm, giving $\left(1+\frac{S}{n}\right)^n < e^S$. So we've actually proved that: $T < e^S$, and putting things together we get

$$1+S < T < e^S$$

So the range of values of T is entirely determined by S. Clearly, if S is finite, T cannot tend to infinity! It's obvious this still holds when we extend n to infinity. QED. By the way: one can check that $e^S$ is a quite good approximation to $T$, with an accuracy far exceeding that of $\left(1+S\right)$, so in many situations we can simply use $T \approx e^S$.

Now let's bring out our "golden key"—$\xi \left(s\right)=\prod\limits_{p} \left(1-p^{-s}\right)^{-1}$. Here we only need the case s=1, and we truncate the right-hand side to primes up to p, i.e.

$$\begin{aligned} &\left(1+\frac{1}{2}+\frac{1}{2^2}+...\right)\left(1+\frac{1}{3}+\frac{1}{3^2}+...\right)...\left(1+\frac{1}{p}+\frac{1}{p^2}+...\right)\\ =&\frac{2}{2-1}\cdot \frac{3}{3-1}\cdot \frac{5}{5-1}\cdot ...\cdot \frac{p}{p-1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{p}+...\\ > &1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{p} \\ > &\ln\left(p+1\right)\end{aligned}$$

Denote the n-th prime by $p_n$; it's not hard to show that $\frac{p_n}{p_n -1} < 1+\frac{1}{p_{n-1}}$.

Hence

$$\begin{aligned}\ln\left(p+1\right) < &\frac{2}{2-1}\cdot \frac{3}{3-1}\cdot \frac{5}{5-1}\cdot ...\cdot \frac{p}{p-1} \\ < &2\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right)...\left(1+\frac{1}{p}\right)\\ < &2e^{\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{p}\right)}\end{aligned}$$

Writing $Q=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{p}$, i.e., $\ln\left(p+1\right) < 2e^Q$, we get

$$Q > \ln \ln \left(p+1\right) -\ln2$$

Since there are infinitely many primes (the number of primes is unbounded and keeps increasing), the sum of the reciprocals of all primes diverges! QED.

Finally, based on the earlier fact that "$e^S$ is a quite good approximation to $T$," we may likewise conclude that $ \ln \ln p$ is a quite good approximation to Q.