Is $\sum_{p}\frac{1}{p!}=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{5!}+...$ irrational? Is there known way to determine whether the infinite sum below is rational or not?
$$\sum_{p}\frac{1}{p!}=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{5!}+...$$

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
$0 < T < 1$.

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.
