# 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

$\{a_n\}$ 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!

• Presumably, you know this is true? Where did you get this result? In what context? – Thomas Andrews May 13 '14 at 3:00
• Hint: If $x = \frac{p}{q}$ is a rational number, then for any positive integer $N$, the fractional part of $Nx$, $\{ Nx \} = Nx - \lfloor Nx \rfloor$, is either $0$ or $\ge \frac{1}{q}$. What can you say about $\{ a_1 a_2 \ldots a_n \sum_{k=1}^\infty \frac{1}{a_k} \}$ for large $n$? – achille hui May 13 '14 at 3:06

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 $\frac12$.