# Rigorously demonstrate the conditions that must be satisfied for an infinite sum of rationals to converge to an irrational [closed]

Rigorously demonstrate the conditions that must be satisfied for an infinite sum of rationals to converge to an irrational.

I hate posting a question with so little forethought, but I really have no idea where to start.

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Nobody knows how to decide this in full generality. It's a nontrivial theorem that $\zeta(2k) = \sum_{n \ge 1} \frac{1}{n^{2k}}$ is irrational (in fact transcendental) for all integers $k \ge 1$, it's a more nontrivial theorem that $\zeta(3)$ is irrational, and I believe it's open whether $\zeta(2k+1)$ is irrational for any particular $k > 1$. –  Qiaochu Yuan Mar 5 '12 at 23:26
@Listing My apologies. –  badreferences Mar 5 '12 at 23:32
@QiaochuYuan Alright, I guess I was asking a bad question, then. Are there any papers on the subject, though? –  badreferences Mar 5 '12 at 23:34
@badreferences: that's still a "bad question." There is literally nothing you can say about an infinite series of rational numbers in full generality because it is possible to encode e.g. the Halting problem in questions about such series. –  Qiaochu Yuan Mar 5 '12 at 23:41
@QiaochuYuan, my understanding is that no one has proved that $\zeta(3)$ is transcendental. –  Gerry Myerson Mar 6 '12 at 0:35

## closed as not a real question by Bruno Joyal, Listing, t.b., Asaf Karagila, Zhen LinMar 31 '12 at 0:03

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Liouville's constant $\sum\limits_{n\geqslant1}10^{-n!}$ is a well known irrational number. Likewise, for every integer sequence $(q_n)_{n\geqslant1}$, the sum of the series $\sum\limits_{n\geqslant1}q_n^{-1}$ is irrational as soon as $(q_n)_{n\geqslant1}$ grows fast enough. Here, fast enough could mean that $q_1\geqslant2$ and that $q_{n+1}\geqslant (q_n)^n$ for every $n\geqslant1$.

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Neat facts, although the original question is the other way around. It asks for necessary conditions for the sum to converge to an irrational. –  Listing Mar 6 '12 at 2:14
@Listing: Indeed... which is the reason why the original question is nearly unanswerable. –  Did Mar 6 '12 at 6:00