 Solution

The Liouville approximation theorem states that an irrational algebraic number $\alpha$ of degree $d\ge 2$ satisfies $|\alpha -a/b|\ge C_{\alpha }{b}^{-d}$ for all integers $a$ and $b\ge 1$, for some $C_{\alpha }>0$. The constant is independent of the rational approximation.

The required condition is infinitely many of the coefficients equal one. If only finitely many do, the series is a finite sum of rational numbers. Conversely let $\alpha$ be the sum with infinitely many nonzero terms. Its truncation $s_N$ has denominator dividing $q_N=10^{N!}$. For $N\ge 2$, its strictly positive tail satisfiesIf $\alpha =a/b$ were rational, the distinct fractions $\alpha ,s_N$ would have distance at least $1/(b{q}_N)$, contradicting the tail bound for large $N$. Thus $\alpha$ is irrational. If it were algebraic of fixed degree $d$, apply the Liouville approximation theorem to $s_N$. Even if its reduced denominator is smaller than $q_N$, the lower bound is at least $C_{\alpha }{q}_N^{-d}$. The displayed upper bound contradicts this as $N\to \infty$. HenceThe coincidence $0!=1!$ changes only the first rational contribution and has no effect on this criterion.