Let $A = \{a_i\}_{i = 1}^{\infty}$ be a sequence of positive integers. If the terms in $A$ grow 'too fast', can we determine that $$S_A = \sum_{i = 1}^{\infty} \frac{1}{a_i}$$ is irrational ? More formally, is there a criteria that says if $a_i = \Omega(f(i)),$ then $S$ is irrational ? If not, for every sequence $A$ such that $S_A$ converges to a rational number, can we find a sequence of positive integers $B = \{b_i\}_{i = 1}^{\infty}$ such that $$\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = 0 \ $$ and $S_B$ also converges to a rational number ?

This question was inspired by the following fact:

$\sum_{k = 1}^{\infty} \frac{1}{(k!)^2}$ and $\sum_{k = 1}^{\infty} \frac{1}{2^{k^2}}$ are both irrational.

  • 3
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Mar 3, 2017 at 2:46
  • $\begingroup$ Well, you can always take a real multiple of $a_n$ that makes the sum rational, so the answer is generally no for the question on the $a_n$. $\endgroup$ Mar 3, 2017 at 2:49
  • $\begingroup$ @Fimpellizieri: what do you mean by a real multiple of $a_n$ ? $\endgroup$ Mar 3, 2017 at 2:49
  • $\begingroup$ If $a_i = \Omega(f(i))$, then $c\cdot a_i = \Omega(f(i))$. Choose $c$ so that $S_{cA}$ is rational. $\endgroup$ Mar 3, 2017 at 3:17
  • $\begingroup$ @Fimpellizieri But then the $a_i$'s are not positive integers $\endgroup$ Mar 3, 2017 at 3:18

3 Answers 3


An example that appears to be at the cusp is Sylvester's sequence:

$$a_0 = 2, \,\,\, a_n = 1 + \prod_{k=0}^{n-1}a_k = 1+ a_{n-1}(a_{n-1} -1),$$

with the closed form $a_n = \lfloor C^{2^{n+1}} + \frac{1}{2} \rfloor$ with $C \approx 1.26$ and a doubly exponential growth rate.

The sum is rational with

$$\sum_{n=0}^\infty \frac{1}{a_n} = 1,$$


$$ \frac{1}{a_{n+1} - 1} = \frac{1}{a_n(a_n - 1)} = \frac{1}{a_n - 1} - \frac{1}{a_n}\\ \implies \frac{1}{a_n} = \frac{1}{a_n - 1} - \frac{1}{a_{n+1} - 1} $$


Although now deleted, the comment to my answer from @charMD cited theorem 3 in 128p of https://www.renyi.hu/~p_erdos/1963-18.pdf

where it is proved that if $a_{n+1} \geqslant 1 + a_n(a_n - 1)$ and strictly greater for infinitely many $n$ then $\sum(1/a_n)$ is irrational.


Adam Hughes essentially answers the question in the comments. Namely, if $a_n>\exp(\prod_{i<n} a_i),$ then by Liouville's lemma, your sum is transcendental (hence, irrational), so doubly exponential growth is sufficient.However, since all you want is irrationality, you can apply the actual statement of Liouville's lemma to remove the $\exp,$ and have $a_n>(\prod_{i<n} a_i)^{(1+\epsilon)},$ for some $\epsilon > 0.$

  • $\begingroup$ But in this example, your $f(i)$ depends on $a_i$ $\endgroup$
    – N. S.
    Mar 5, 2017 at 4:58
  • $\begingroup$ @N.S. Actually, no, it does not. If the sequence grows fast enough that condition is satisfied. $\endgroup$
    – Igor Rivin
    Mar 5, 2017 at 16:01
  • $\begingroup$ Nope, is not true. No matter how fast $f(i)$ is growing, you can always make your condition fail by picking some $a_n$ to be much bigger than $f(n)$. For example, if you pick $a_1$ bigger than $f(2017)$ and $a_j=f(j)$ up to $2016$, your condition clearly fails for $n=1,..., 2016$. Then repeat. Or something like $a_{2^n}>f(2^{n+1})$ and $a_j=f(j)$ otherwise. $\endgroup$
    – N. S.
    Mar 5, 2017 at 16:32

If a number is rational, say $\frac{p}{q}$, then $\left|\frac{p}{q}-\frac{m}{n}\right|\geq\frac{1}{nq}$. So the difference from any test rational $\frac{m}{n}$ times the denominator $n$ has to be bounded below unless you hit your number spot on. If we find an increasing sequence of rational numbers for which that is not the case, then the limit is irrational.

So assuming the sequence grows as fast as a geometric series (that's a reasonable enough assumption of growth) so that the sum can be estimated by the next term, if you have $\frac{\text{lcm}(a_1, \ldots a_n)}{a_{n+1}}\rightarrow0$ as $n\rightarrow\infty$, then the sum $\sum_{i=1}^{\infty}\frac{1}{a_i}$ is irrational.

These cover the two cases in your question, though neither is doubly exponential. It even shows that $e$ is irrational. But it doesn't quite answer the question. It does, however, provide a set of $f(i)$'s that are good enough. Specifically, we can let $f(i)$ be a sequence for which $f(i)^{1/2^i}$ increases to $\infty$. Because if we have a sequence of $a_i>f(i)$, then for any $\epsilon$ we take the smallest $i$ for which $a_i>\left(\frac{1}{\epsilon}\right)^{2^{i-1}}$ which exists; then

$\frac{\text{lcm}(a_1,a_2,\ldots a_{i-1})}{a_i}<\frac{a_1a_2\ldots a_{i-1}}{a_i}<\frac{\left(\frac{1}{\epsilon}\right)^1\left(\frac{1}{\epsilon}\right)^2\ldots\left(\frac{1}{\epsilon}\right)^{2^{i-2}}}{\left(\frac{1}{\epsilon}\right)^{2^{i-1}}}=\epsilon$

And this is good enough, because we only need a subsequence of the above to tend to 0, not the entire sequence. So for any sequence $u(i)$ increasing to infinity, $u(i)^{2^i}$ works.

  • $\begingroup$ But your $f(n)$ depends on $a_n$. $\endgroup$
    – N. S.
    Mar 5, 2017 at 16:33

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