Is $\sum_{n \ge 1}{\frac{p_n}{n!}}$ irrational, where $p_n$ is the $n^{\text{th}}$ prime number?

This question is spurred by the comment thread on this question where I presented a rough idea of a proof similar to the well-known proof that $e$ is irrational. I will try to post my idea as a self-answer, other attempts are welcome too of course.

EDIT: Is there a way to prove this using the prime number theorem only?

  • 2
    $\begingroup$ What is $p_n$ ? $\endgroup$ – Archis Welankar Jan 9 '16 at 4:53
  • $\begingroup$ The $n^{\text{th}}$ prime number. $\endgroup$ – Dan Brumleve Jan 9 '16 at 4:53

We know from Ingham's 1937 result that $p_{n+1}-p_n = O(p_n^{0.7})$, so for sufficiently large $n$:

$p_{n+1} - p_n \lt p_n^{0.8} \lt \frac{n}{3} \lt \frac{n}{3} + \frac{p_{n+1}}{n+1}$

where the middle inequality is a consequence of the prime number theorem. We can rewrite this as:

$\frac{p_{n+1}}{n+1} - \frac{p_n}{n} \lt \frac{1}{3}$

From this we can conclude that for infinitely many $n$, the fractional part of $\frac{p_n}{n}$ is less than $\frac{1}{2}$, since it is unbounded (again by the prime number theorem) and it can't jump by more than $\frac{1}{3}$ in one step.

Next, suppose $\sum_{i \ge 1}{\frac{p_i}{i!}} = \frac{a}{b}$. Clearly, $(n-1)! \cdot \sum_{1 \le i \lt n}{\frac{p_i}{i!}}$ is an integer, and if we require $n \gt b$, then $(n-1)! \cdot \sum_{n \le i}{\frac{p_i}{i!}}$ must also be an integer. But as shown above, $n$ can be chosen so that the first term of the latter sum, $\frac{p_n}{n}$, has a fractional part less than $\frac{1}{2}$; and the sum of the following terms $\frac{p_{n+1}}{n \cdot (n+1)} + \frac{p_{n+2}}{n \cdot (n+1) \cdot (n+2)} + \dots$ will also be less than $\frac{1}{2}$ provided $n$ is large enough (again using PNT). This contradicts our assumption that the number is rational.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.