$\{a_n\}$ is a strictly increasing sequence of positive integers such that $$\lim_{n\to\infty}\frac{a_{n+1}}{ a_n}=+\infty$$ Can one conclude that $\sum\limits_{n=1}^\infty\frac1{a_n}$ is an irrational number? a transcendental number?

A special case is $a_n=n!$, $e$ is a transcendental number.

Another special example is Liouville number $\sum\limits_{n=1}^\infty\dfrac1{10^{n!}}$ is a transcendental number, too.

so the question, if true, may be difficult.

The question is a generalization of If $(a_n)$ is increasing and $\lim_{n\to\infty}\frac{a_{n+1}}{a_1\dotsb a_n}=+\infty$ then $\sum\limits_{n=1}^\infty\frac1{a_n}$ is irrational

  • $\begingroup$ I have heard about specific classes of sequences that converge faster than a geometric series and always have irrational limits, but I cannot remember the exact hypothesis. Also, if we allow the $a_n$ to be non-integers, it's clearly not true. $\endgroup$
    – Arthur
    May 13, 2014 at 15:39
  • $\begingroup$ @ᛥᛥᛥ Whoops, I missed that! $\endgroup$ May 13, 2014 at 15:46
  • $\begingroup$ In the case or $e$, you have $a_n\mid a_{n+1}$ for every $n$. ${}\qquad{}$ $\endgroup$ May 13, 2014 at 16:05

2 Answers 2


The conjecture is false.

Take $S_0 = 0$, and for $n=1,2,\cdots$ define $a_n$ to be the least integer such that $S_{n-1}+1/a_n < 1$. Then $\sum_{n=1}^\infty 1/a_n = 1$ but $a_{n+1}/a_n$ grows very fast... $$ a_1 = 2\\ a_2 = 3\\ a_3 = 7\\ a_4 = 43\\ a_5 = 1807\\ a_6 = 3263443\\ a_7 = 10650056950807\\ a_8 = 113423713055421844361000443\\ a_9 = 12864938683278671740537145998360961546653259485195807 $$


The answer is NO.

Consider the Sylvester's sequence (OEIS A000058):

$$(s_0, s_1, \ldots ) = ( 2, 3, 7, 43, 1807, 3263443, 10650056950807, \ldots)$$ defined recursively by the relation

$$s_n = \begin{cases} 2,& n = 0,\\ s_{n-1}(s_{n-1}-1)+1,& n > 0 \end{cases}$$

It is known that its reciprocals give an infinite Egyptian fraction representation of number one:

$$1 = \frac12 + \frac13 + \frac17 + \frac{1}{43} + \frac{1}{1807} + \cdots$$

It is also easy to check $\displaystyle\;\lim_{k\to\infty} \frac{s_{k+1}}{s_k} = \infty\;$. If you set $a_n = s_{n-1}$ for $n \in \mathbb{Z}_{+}$, you get a counterexample of what you want to show. i.e $\displaystyle\;\sum_{n=1}^\infty \frac{1}{a_n}\;$ need not be irrational.

  • $\begingroup$ To me it seems clearer to state the definition by saying $s_{n+1} = 1 + \prod_{k=1}^n s_n$. ${}\qquad{}$ $\endgroup$ May 13, 2014 at 16:19
  • $\begingroup$ @MichaelHardy that might be true. However, I start from constructing the infinite Egyptian fraction myself, come up the recursion relation I wrote down before I locate the name and other properties of this sequence. $\endgroup$ May 13, 2014 at 16:26

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