# 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:

However, the answer key says that this is not a rational number. Could anyone help me understand why this is not a rational number?

• The sum of a finite series of rational numbers is indeed rational. But, an infinite series of rational numbers may or may not be rational. – Doug M Aug 24 '16 at 17:22
• Neither the numerator nor the denominator are rational. The sum of two rational numbers is a rational, and induction shows that any finite sum of rational numbers is rational, but it's entirely possible for the infinite sum of rational numbers to be irrational. – Akiva Weinberger Aug 24 '16 at 17:23
• Numerator is not exactly $\pi/4$. It converges to $\pi/4$. – absolute friend Aug 24 '16 at 17:24
• @absolutefriend: You just contradicted yourself. Infinite sums are defined to be the limit of their partial sums. – Deusovi Aug 24 '16 at 18:23
• @fleablood: that argument doesn't work. After all, $$1=\frac{1+1/4+1/9+\cdots}{1+1/4+1/9+\cdots}.$$ – Martin Argerami Aug 24 '16 at 21:04

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.

• ... and all of the factors other than $\pi$ are integers. – user14972 Aug 25 '16 at 4:42
• @Hurkyl I do acknowledge that the other factors need to be integers. I just left it to the reduced form to say it implicitly. $\ddot\smile$ – Parcly Taxel Aug 25 '16 at 16:34
• Proof that π itself is irrational is left as an exercise for the reader. – Dan Aug 25 '16 at 21:45

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.

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 don't like the proof of Leibniz' formula via $\arctan 1$, because the series does not (necessarily) converge at $1$ (which is the radius of convergence of the series). I think there is a more involved proof that this point is convergent to the power series using Abel's theorem, but it is not at all as simple as most books make it out to be. – Mario Carneiro Aug 24 '16 at 18:50
• @MarioCarneiro : Yes, Abel's theorem is another "elided complication". $\qquad$ – Michael Hardy Aug 24 '16 at 19:23

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.

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.

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.

• Thanks. Your answer was the easiest to understand .. :) – user148849 Aug 24 '16 at 17:53
• Thanks, I didn't like the answers that were jumping to the place that you were supposed to recognize the series. It didn't answer your basic questions. – Doug M Aug 24 '16 at 18:24
• Your argument via LCMs is incorrect. Note that $\sum_{n=1}^\infty(\frac1n-\frac1{n+1})=1$ but the denominators of the rational numbers of the sum are $n(n+1)$, and the LCM of these over all $n$ contains every prime number and hence is infinite... yet the infinite sum is an integer. I don't know any easy proof that $\pi$ is irrational such as this one. – Mario Carneiro Aug 24 '16 at 18:44
• You should address the issue pointed out by Mario above. As written the second half of the answer is dead wrong. – Kibble Aug 25 '16 at 14:30

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