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,