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

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