Test of being a rational number for $(1-\frac13+\frac15-\frac17+\cdots)/(1+\frac14+\frac19+\frac1{16}+\cdots)$
The numerator is the Leibniz/Gregory series, which sums to $\frac\pi4$. The denominator is the subject of the famous Basel problem, which Euler worked out as $\frac{\pi^2}6$. If we use these results in the fraction: $$F=\frac{\frac\pi4}{\frac{\pi^2}6}=\frac{3}{2\pi}$$ which is an irrational number because there remains a $\pi$ in it.
My thoughts: Sum and product of two rational numbers is a rational number.
Difference of two rational numbers is a rational number.
Division of two rational numbers should also be a rational number. (Denominator is not zero)
These only apply to finite number of operands.
It is easy to see that they don't apply in the infinite case. Consider for example any $x \in [0, 1)$. Then
$$ x = \sum_{k = 1}^{\infty} \frac{n_k}{10^k} $$
where $n_k$ is the $k^{th}$ decimal digit of $x$. So any $x \in [0,1)$ can be expressed as a (possibly infinite) sum of rationals, but of course not every $x \in [0,1)$ is a rational.
I am going to keep this part that I had initially put forward as a hint.
$$1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\ldots=\sum_{n=1}^\infty (-1)^n\frac{1}{2n+1}=\frac{\pi}{4}$$ $$1+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\ldots=\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$
To add more to this answer, I would like to state that these are well known sequences, as already mentioned by Parcly Taxel. I would only make one point clear, sum of an infinite series of rational numbers does not always put up a rational number as the sum. The first series is just the Taylor expansion of $\arctan x$ about $x=0$ and the second one, well I remember deriving the result from the Fourier series representation of $x^2$. It can also be derived as per Euler. Always keep in mind that having an irrational number in your mathematical expression neither confirms nor nullifies the chance of your expression summing down to an irrational or rational number. It solely and only depends on the sum.