Solution
= Solution
Let
$$
x=\sum_{j=1}^\infty\frac{\theta_j}{j!}
$$
and suppose $x=p/q$. Choose $n\geq\max(q,2)$ such that some $\theta_j=1$ for $j>n$. Then $n!x$ is an integer, while its first $n$ terms also sum to an integer. Their difference is strictly positive and satisfies
$$
0<n!\sum_{j>n}\frac{\theta_j}{j!}
\leq\sum_{r=1}^\infty\frac1{(n+1)^r}<1.
$$
This cannot be the difference of two integers, so $x$ is irrational.