Test of being a rational number for $(1-\frac13+\frac15-\frac17+\cdots)/(1+\frac14+\frac19+\frac1{16}+\cdots)$ Is the following expression a rational number?
$$\frac{1-\dfrac13+\dfrac15-\dfrac17+\cdots}{1+\dfrac14+\dfrac19+\dfrac1{16}+\cdots}$$
My thoughts:

*

*The sum and product of two rational numbers is a rational number.

*So is the difference.

*As well as the quotient, when the denominator is not zero.
However, the answer key says that this is not a rational number. Could anyone help me understand why this is not a rational number?
 A: 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.
A: 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.
A: 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.
Does this prove that the sum is irrational?  Unfortunately, no.  But it should set off signals that I might be.  And certainly as we start adding partial sums, we see a fraction that is becoming increasingly unwieldy. 
So, we have an irrational number / irrational number.  It is possible with only that information that the quotient is rational, but it is unlikely.   
A: $$=\frac{\tan^{-1} 1}{\zeta (2)}=\frac{\pi/4}{\pi^2/6}$$
A: 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.
A: 
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.
