Convergent rational series: which ones remain rational? 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.
 A: 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,
A: 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. 
