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Due to closure under addition, it is obvious that a finite sum of rationals is rational. The infinite ones, however (assuming they don't diverge), may remain rational, such as $\sum_{n \in \mathbb{N}} 2^{-n}$, or not, like $\sum_{n \in \mathbb{N}} n^{-2}$.

Is there a criterion to find out whether a (convergent) series of rationals is rational or irrational without calculating it?

P. S.: insights on the analogous question with infinite products are welcome.

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There can't be a known foolproof way to do this, because you can express $\pi + e$ as an (explicitly given) convergent sequence of rationals, and AFAIK it is an open problem whether $\pi + e$ is irrational. – Rahul Jun 11 '11 at 17:34
Nope. This is very hard. Incidentally, there's a fun trick here: one way to show that a set $A$ is infinite is to show that $\sum_{a \in A} f(a)$ or $\prod_{a \in A} f(a)$ is irrational for some function $f : A \to \mathbb{Q}$. For example, since $\prod_{p \text{prime}} \left( \frac{1}{1 - p^{-2}} \right) = \frac{\pi^2}{6}$ is irrational, there are infinitely many primes. – Qiaochu Yuan Jun 11 '11 at 17:37
up vote 5 down vote accepted

As mentionned in the comments, there is no general solution to this problem. A simple example is the series $$\zeta(5)=\sum_{n=1}^\infty n^{-5}$$ which has not yet been proven to be rational or irrational.

However, sometimes we can say that a particular series must be transcendental if it can be "very well approximated" by rational numbers.

Irrationality Measure: For a real number $x$, consider $$E(x)=\left\{ \alpha\in\mathbb{R}:\ \text{there exists infinitely many }q\ \text{with}\ \biggr|x-\frac{p}{q}\biggr|<\frac{1}{q^{\alpha}}\right\}.$$ Let $\mu(x)=\sup\left(E(x)\right).$ If $x$ is rational, then $\mu(x)=1$, and if $x$ is a quadratic irrational, then by using some theorems in continued fractions we know that $\mu(x)=2$. The Thue-Siegel-Roth Theorem tells us more generally that if x is algebraic, and not rational, then $\mu(x)=2$. Unfortunately this does not give a complete characterization since $e$ is transcendental, and $\mu(e)=2$.

Series which are transcendental: Consider the following series where $q,a$ are integers: $$\alpha_{q}(a)=\sum_{n=1}^{\infty}\frac{1}{q^{a^{n}}}.$$Then if $a\geq3$ we know that this must be transcendental. If $a=2$, this test will not tell us, since then $\mu(\alpha_q(a))=2$. But this does mean that when $a=2$, the series is irrational. We can apply these ideas to certain series which have terms decreasing fast enough. Another example is $$c=\sum_{n=1}^{\infty}10^{-n!}.$$ This is called Liouvilles constant, and was one of the first examples of a transcendental number. Since $\mu(c)=\infty$, it follows that $c$ is transcendental.

Hope that helps,

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On the first topic, you don't clarify what kind of number $p$ and $q$ are. I assume they're positive integers. – Luke Jun 12 '11 at 4:53

In fact one can reduce a very general class of problems to the question of determining whether a certain series of rational numbers converges to a rational number or not. Let $A$ be a subset of the natural numbers (for example, the set of twin primes). Then $A$ is infinite if and only if

$$\sum_{a \in A} 2^{-a^2}$$

is irrational, since if $A$ is infinite then the base-$2$ expansion of the above number cannot be eventually periodic. Many, many arbitrarily hard problems can be encoded as the problem of determining whether a set of natural numbers is infinite; for example, it is undecidable whether the set of natural numbers a Turing machine recognizes is infinite by Rice's theorem.

In other words, what often doesn't come through in a calculus course is that the only series you've ever seen summed are likely to be much simpler than a "generic" series, which can be arbitrarily complicated. So one has to place strong restrictions on what kind of series are considered to have any hope of saying something general.

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Out of curiosity, do you know of any problems regarding the infinitude of some set which are solved more easily when rephrased as a question about the irrationality of some series? The infinitude of the primes example above is pretty good, but usually proving that $\zeta(2)=\frac{\pi^2}{6}$ is much "harder" then just using Euclids proof. (I hope the wording of this question makes sense) – Eric Naslund Jun 11 '11 at 22:51
@Eric: no I do not. I just happen to like that example; it uses the Riemann zeta function but not in the usual way. – Qiaochu Yuan Jun 11 '11 at 22:53

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