Beautiful proofs(#2): Euler’s Sum

1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots = \frac{\pi^2}{6}

Say what? This one blew my mind when I first encountered it. But it turns out Euler was the one who came up with it and it’s proof is just beautiful!

Prerequisite
Say you have a quadratic equation f(x) whose roots are r_1,r_2 , then you can write f(x) as follows:

f(x) = (x-r_1)(x-r_2) =  0   (or)

f(x) = (r_1-x)(r_2-x) =  0   (or)

f(x) =  (1- \frac{x}{r_1})(1- \frac{x}{r_2}) =  0

f(x) = 1 - (\frac{1}{r_1} + \frac{1}{r_2}) + \frac{x^2}{r_1 r_2} = 0

As for as this proof is concerned we are only worried about the coefficient of x , which you can prove that for a n-degree polynomial is:

a_1 = - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})

where r_1,r_2 \hdots r_n are the n-roots of the polynomial.

 

Now begins the proof

It was known to Euler that

f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots

But this could also be written in terms of the roots of the equation as:

f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})y + \hdots

Now what are the roots of f(y) ?. Well, f(y) = 0 when \sqrt{y} = n \pi i.e y = n^2 \pi^2 *

The roots of the equation are y = \pi^2, 4 \pi^2, 9 \pi^2, \hdots

Therefore,

f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots = 1 -( \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \hdots )y + \hdots

Comparing the coefficient of y on both sides of the equation we get that:

\frac{1}{6} = \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \frac{1}{ 9 \pi^2} + \hdots

\zeta(2) = \frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \hdots 

Q.E.D

* n=0 is not a root since
\frac{sin(\sqrt{y})}{\sqrt{y}} = 1 at y = 0

** Now if all that made sense but you are still thinking : Why on earth did Euler use this particular form of the polynomial for this problem, read the first three pages of this article. (It has to do with convergence)

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