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$\{a_n\}$ 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 $\sum\limits_{n=1}^\infty\frac1{a_n}$ is irrational

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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. –  Arthur May 13 at 15:39
    
@ᛥᛥᛥ Whoops, I missed that! –  Fly by Night May 13 at 15:46
    
In the case or $e$, you have $a_n\mid a_{n+1}$ for every $n$. ${}\qquad{}$ –  Michael Hardy May 13 at 16:05

3 Answers 3

up vote 6 down vote accepted

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.

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To me it seems clearer to state the definition by saying $s_{n+1} = 1 + \prod_{k=1}^n s_n$. ${}\qquad{}$ –  Michael Hardy May 13 at 16:19
    
@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. –  achille hui May 13 at 16:26

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... $$ a_1 = 2\\ a_2 = 3\\ a_3 = 7\\ a_4 = 43\\ a_5 = 1807\\ a_6 = 3263443\\ a_7 = 10650056950807\\ a_8 = 113423713055421844361000443\\ a_9 = 12864938683278671740537145998360961546653259485195807 $$

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edAssume the sum is rational $p/q$.

Then $$p = \sum_{n=1}^{m}\frac{q}{a_n} + \sum_{n=m+1}^{\infty}\frac{q}{a_n}$$

and

$$qa_1\ldots a_m\sum_{n=m+1}^{\infty}\frac{1}{a_n} < qa_1\ldots a_m\sum_{n=m+1}^{\infty}r^{-n}$$

$$pa_1\ldots a_m - \sum_{n=1}^{m}qa_1\ldots a_{n-1}a_{n+1}\ldots a_m = qa_1\ldots a_m\sum_{n=m+1}^{\infty}\frac{1}{a_n} \tag{*}$$

From the limit we know for $n$ sufficiently large, ${1/a_n}$ is dominated by the terms of a convergent goemetric series so

$$qa_1\ldots a_m\sum_{n=m+1}^{\infty}\frac{1}{a_n} < qa_1\ldots a_m\sum_{n=m+1}^{\infty}r^{-n}$$

and $$qa_1\ldots a_m\sum_{n=m+1}^{\infty}\frac{1}{a_n} < qa_1\ldots a_m\frac{r^m}{1-r}$$

We can choose $r$ sufficiently small and and $m$ sufficiently large so the RHS of (*) is less than $1$ and the LHS is an integer, a contradiction

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The proof is wrong, because first, you fix $r>0$, and $r$ depends on $m$. Intuitively, the smaller is $r$, the greater must be $m$. You cannot guarantee that $qa_1\ldots a_m\frac{r^m}{1-r}$ is less than $1$, because it is involved a product of $m$ integers, and by the way, no smaller integers at all... –  ajotatxe May 13 at 16:55
    
Thanks -- I threw this together in a hurry as everyone jumped on this problem. I see now where it breaks down. –  RRL May 13 at 19:18
    
"I threw this together in a hurry as everyone jumped on this problem" This must be one of the most absurd motives to "throw something together in a hurry" on a maths site. –  Did May 15 at 17:48

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