$\{a_n\}$ 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!

  • $\begingroup$ Presumably, you know this is true? Where did you get this result? In what context? $\endgroup$ May 13, 2014 at 3:00
  • 2
    $\begingroup$ 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 $n$? $\endgroup$ May 13, 2014 at 3:06

1 Answer 1


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 $\frac12$.


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