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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!)

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

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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}

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