Test of being a rational number

The sum of a 2 rational numbers is rational

The sum of a finite number of rational numbers is rational.

But, an infinite series of rational numbers may or may not be rational.

Can't we just keep adding 2 at a time and keep going on like that?

No.

Here is another example, that perhaps will help you get your head around it.

I hope we can agree that $\pi$ is irrational. and that the first few digits of the decimal expansion of $\pi$ is $3.14159$ but $3.14159$ is rational. In fact, any finite expansion of $\pi$ is rational. We can add more digits.

$3.14159 + 0.0000026 = 3.1415926$

And adding digits is a summation of rational numbers.

But it is only when we accept that it is the infinite non-repeating decimal that we we have the irrational number that is $\pi$

If not, how do I test?

Let's look at the numerator.

$1-\frac 13 + \frac 15 - \frac 17+ \cdots$

Now, you might recognize this as the Taylor expansion of $\tan^{-1} 1$

But you might not.

If it is rational then there exists integer $p,q$ such that $\frac pq = \sum \frac {(-1)^n}{2n+1}.$

If this is going to sum up to a single fraction, what is the common denominator?

It is $lcm (3,5,7,9,11\cdots)$ Since we have every odd number, we have every prime number (other than 2), and the lcm is infinite.

There is no $p,q.$ such that $p/q = \frac {\pi}{4}$

The numerator is irrational.

So, we have an irrational number / irrational number. It is possible with only that information that the quotient is rational, but it is unlikely.

If operations on TWO rational numbers yields a rational number, then those same operations on FINITELY MANY rational numbers yields a rational number. You can prove that by mathematical induction. But that doesn't apply to a sum of infinitely many numbers. (If it did, then you could prove by mathematical induction that every countably infinite set is finite.)

$$ \frac d {dx} \left( x - \frac {x^3} 3 + \frac{x^5} 5 - \frac {x^7} 7 + \cdots\right) = 1 - x^2 + x^4 - x^6 + \cdots = \frac 1 {1+x^2} $$ (since you're summing a geometric series), and therefore $$ x - \frac {x^3} 3 + \frac{x^5} 5 - \frac {x^7} 7 + \cdots = \arctan x, $$ so if $x=1$ you get the series in your numerator, which is therefore $(\arctan 1) = \dfrac\pi4.$

The series in the denominator is much harder to sum, but it should come to $\dfrac{\pi^2} 6.$

I've elided some details: in particular, the derivative of a sum of TWO functions is the sum of the derivatives, and therefore the same is true of the sum of FINITELY many, but what about infinitely many? Sometimes it doesn't work there, so there's some theory of power series to deal with.

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.

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.

$$=\frac{\tan^{-1} 1}{\zeta (2)}=\frac{\pi/4}{\pi^2/6}$$

If it is rational, you will be able to find a common denominator. The common denominator has to be the least common multiple of every single natural number. This is impossible.

So it is irrational.

The "catch" that the sum and difference of rational numbers is rational is that it is the sum and difference of a FINITE number of rational numbers that is rational; this is the sum and difference of an infinite number of rational numbers. That doesn't have to be rational.

An intuitive example of how it's possible for an infinite sum of rational numbers can have a result that is not rational is to consider any irrational number in it's decimal expansion.

e.g. $\pi = 3.1415926...$

This is $3 + .1 + .04 + .001 + .... $ etc. This is an infinite sum of rational numbers resulting in an irrational result.

Of course this answer simply begs the real question. We have to define how irrational numbers can actually exist and why we think decimal expansions are valid and what are real numbers. But I'll leave that to your analysis professor...

Oh, all right. One thing... An irrational number can not be express as a rational number. Be it "is infinitely close to rational numbers". What does that mean? It means if we find a rational number close to it we can add or subtract another rational number to get even closer to it. And that means the irrational number is itself an infinite sum of rational numbers. All irrational numbers are.

So there is no reason to assume your expression is rational.

Showing that your expression is related to $\pi$ is a neat trick but very hard to prove.