Here I will discuss another of my favourite formulas:
It boggles the mind how the sum of reciprocals of squared natural numbers could possibly be related to ! Interestingly, the reciprocal of this number, , is the probability that two randomly selected positive integers are relatively prime. That is, choose two integers uniformly and independently between 1 and N, then as N tends to infinity the probability they share no common factor other than 1 is .
Finding the infinite sum was first posed by Pietro Mengoli in 1644 and is now known as the Basel problem. It stumped the great mathematicians of the next ninety years, ranging from Leibniz to Jacob Bernoulli. Finally in 1735 Euler became famous after being the first to find the answer, aged 28. He used his immense talent in numerical analysis (using what is now known as the Euler-Maclaurin formula) to conjecture the sum to be , then this motivated him to use series formulas for to prove the result. We will outline three other proofs below, more proofs can be found in Robin Chapman’s collection of 14 proofs here.
Proof 1: Elementary proof, found on MathOverflow website here.
The idea is to sandwich the sum between two expressions which each converge to as .
We start with the inequality
, where .
This can be seen by drawing a unit circle and comparing the areas of a sector with internal angle radians (area ) , a triangle with unit sides and internal angle (area ) and a right angle triangle with base length 1 and angle (area ).
Next, let . Then from the above inequality,
Now let us evaluate the right side of this inequality, denoted by . We sum terms symmetric about and find
This means we have the recurrence relation with the initial condition (initial terms 2, 10, 42, 170, …). Using mathematical induction or generating functions (z-transforms), one can show that this has the closed form solution . Similarly,
Then from (*),
We see that both left and right sides tend to as , and we conclude that .
Proof 2: Integration by clever change of variables, due to Calabi, Beukers and Kock.
Using the result ,
Then make the remarkable substitution
We find that this sets up a bijective transformation between the unit square and the triangle
(Note that for implies and so or . The inverse map is .)
Also we find
, where is defined above, and has area . Hence
from which .
Proof 3: Proof by Fourier series
This is the proof seen in many textbooks, but requires a more advanced result than the previous two, namely Parseval’s theorem. We use the fact that the complex exponentials where form a complete orthonormal basis in the set of square-integrable complex-valued functions on the unit interval [0,1]. This means such a function on the interval [0,1] can be written as
where are Fourier coefficients which may be found by taking the inner product of both sides of the above with :
In linear algebra terms, we are simply projecting onto the basis vector to find its component in that direction. Parseval’s theorem is simply a generalisation of Pythagoras’s theorem, stating that the square of the norm of is the sum of the squared magnitudes of its components:
We now calculate this for , noting that .
- For ,
Combining these results in Parseval’s theorem gives
from which , as desired.