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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?

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    $\begingroup$ 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
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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.

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