TITLE: 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
QUESTION [11 upvotes]: $\{a_n\}$ is a  strictly increasing sequence of positive integers such that 
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_1a_2\dotsb a_n}=+\infty$$
then $\sum\limits_{n=1}^\infty\frac1{a_n}$ is an irrational number


proof by contradiction?  $\sum\limits_{n=1}^\infty\frac1{a_n}=\frac pq$,
I also try the series $\sum\limits_{n=1}^\infty\frac{x^n}{a_n}$, without any progress. The classical proof of  irrational of $e$ use Taylor formula
The problem was proposed by Jose Luis Daz-Barrero. If $\lim\limits_{n\to\infty}\frac{a_{n+1}}{ a_n}=+\infty$, then $\sum\limits_{n=1}^\infty\frac1{a_n}$ is irrational, too?
Thanks a lot!

REPLY [7 votes]: Suppose towards contradiction that
$$
\sum_{n=1}^\infty\frac1{a_n}=\frac pq \quad \quad p, q \in \mathbb{Z}^+
$$
Let $M = 2qe$.
Then choose $N$ such that for $n > N$, $a_{n} > M a_1 a_2 \ldots a_{n-1}$.
Multiplying the above by $q a_1 a_2 \ldots a_N$, we have
$$
pa_1 a_2 \ldots a_N
= \sum_{n=1}^N q a_1 a_2 \ldots a_{n-1} a_{n+1} \ldots a_N
+ \sum_{n=N+1}^\infty \frac{qa_1a_2\ldots a_N}{a_n}
$$
implying
$$
\sum_{n=N+1}^\infty\frac{qa_1a_2\ldots a_N}{a_n} \in \mathbb{Z}
$$
But then
$$
\frac{qa_1a_2\ldots a_N}{a_n} < \frac{q a_1 a_2 \ldots a_N}{M a_1 a_2 \ldots a_{n-1}}
$$
so
\begin{align*}
0
&< \sum_{n=N+1}^\infty\frac{qa_1a_2\ldots a_N}{a_n} \\
&< \sum_{n=N+1}^\infty \frac{q}{Ma_{N+1}a_{N+2} \ldots a_{n-1}} \\
&< \frac{q}{M} \sum_{n=N+1}^\infty \frac{1}{(n - N - 1)!}
\quad \quad \text{(since } a_n \text{ is increasing)}
 \\
&= \frac{q}{M}e = \frac{1}{2} \\
\end{align*}
which is a contradiction, as there is no integer between $0$ and $\frac12$.