Is $\sum_{p}\frac{1}{p!}=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{5!}+...$ irrational?

Question

Is there known way to determine whether the infinite sum below is rational or not?

$$\sum_{p}\frac{1}{p!}=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{5!}+...$$

Answer 1

Note: I have added a failed attempt to prove that $x$ is transcendental by showing that $x$ is a Liouville number (see https://en.wikipedia.org/wiki/Liouville_number).

Here is a formal proof of Winther's comment:

Let $x =\sum_{p \in P} \frac1{p!} $, where $P$ is a sequence of strictly increasing positive integers.

If $x = \frac{a}{b}$,

$\begin{array}\\ b!x &=b!\sum_{p \in P} \frac1{p!}\\ &=b!\left(\sum_{p \in P, p \le b} \frac1{p!}+\sum_{p \in P, p > b} \frac1{p!}\right)\\ &=b!\sum_{p \in P, p \le b} \frac1{p!}+b!\sum_{p \in P, p > b} \frac1{p!}\\ &= S+T\\ \end{array} $

where $S$ is an integer and

$\begin{array}\\ T &=b!\sum_{p \in P, p > b} \frac1{p!}\\ &\le b!\sum_{n=b+1}^{\infty} \frac1{n!}\\ &= \sum_{n=b+1}^{\infty} \frac1{\prod_{k=b+1}^n k}\\ &< \sum_{n=b+1}^{\infty} \frac1{(b+1)^{n-b}}\\ &\le \sum_{n=1}^{\infty} \frac1{(b+1)^{n}}\\ &=\frac{\frac1{b+1}}{1-\frac1{b+1}}\\ &=\frac1{b}\\ &\le 1\\ \end{array} $

This is a contradiction since $b!x$ and $S$ are integers and $0 < T < 1$.

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Here is the start of a failed attempt to prove that $x$ is transcendental by showing that $x$ is a Liouville number (see https://en.wikipedia.org/wiki/Liouville_number).

Let $(p_n)_{n=1}^{\infty}$ be the sequence, and suppose that $\sup(p_{n+1}-p_n) =\infty $. The primes satisfy this.

Let $r_n =\frac{a_n}{b_n} =\sum_{k=1}^n \frac1{p_n!} $, so that $b_n \le p_n!$.

If $p_{n+1}-p_n =m $, then $p_{n+j} \ge p_n+m+j-1 $ for $j \ge 1$, so

$\begin{array}\\ b_n(x-r_n) &=b_n\sum_{k=n+1}^{\infty} \frac1{p_k!}\\ &\le p_n!\sum_{k=n+1}^{\infty} \frac1{p_k!}\\ &= \frac{p_n!}{p_{n+1}!}\sum_{k=n+1}^{\infty} \frac{p_{n+1}!}{p_k!}\\ &= \frac{p_n!}{p_{n+1}!}\left(1+\sum_{k=n+2}^{\infty} \frac{p_{n+1}!}{p_k!}\right)\\ &<2 \frac{p_n!}{p_{n+1}!}\\ &< \frac{2}{(p_n+1)^m}\\ \end{array} $

so $x-r_n \lt \frac{2}{b_n(p_n+1)^m} $.

However, we need $b_n^m$ in the denominator, not $(p_n+1)^m$. So this does not prove trancendentality.