TITLE: If $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = 0$, can $\sum_0^\infty a_n$ be rational?
QUESTION [9 upvotes]: If a nonzero sequence of rationals
$$a_0, a_1 \dots a_n$$
"decays fast" in the sense that $\lim_{n \rightarrow \infty} a_{n+1}/a_n = 0$, can the series converge to a rational number? That is, can $\sum_0^\infty a_n$ be rational?
The motivation is the idea that $e$ is irrational in some sense because its series $\sum \frac{1}{n!}$ grows too slowly. I was wondering if there was a way to make this precise.
(This is a restatement of this question, with thanks to Barry Cipra!)

REPLY [3 votes]: Second attempt at a solution:
Yes, example:
\begin{eqnarray} a_{n} = \frac{1}{(2)_{n}} - \frac{1}{n!} \end{eqnarray}
where the Pochhammer symbol is $(\beta)_{n} = \beta(\beta+1) \cdots (\beta+n-1)$. Then,
\begin{eqnarray} \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=0 \text{ and } \sum_{n=0}^{\infty} a_{n} = -1. \end{eqnarray}