If $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = 0$, can $\sum_0^\infty a_n$ be rational?

How about something like $.1, .011, .000111, .0000001111, \dots$? Clearly the sum is $.11111... = 1/9$ and the $n$th term is on the order of $10^{-n}$ times the $(n-1)$st term.

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}