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

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!)

• I may be oversimplifying, but what about an ordinary convergent geometric series? – imranfat Apr 21 '16 at 14:05
• @imranfat The ratio is constant. – Clement C. Apr 21 '16 at 14:06
• Incidentally, the "grows slowly" should apply to the series, not the sequence. The sequences decays very fast with this assumption. – Clement C. Apr 21 '16 at 14:07
• @ClementC.: Thanks, you're right. Edited. – Eli Rose -- REINSTATE MONICA Apr 21 '16 at 14:08
• Well, I guess I was not understanding the "grows slowly" wording in the question then... – imranfat Apr 21 '16 at 14:09

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.
\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,