Let q = ζ(2). If we consider only the even terms of the series, 1/4 + 1/16 + 1/36 + …, the result is ¼q. Thus the odd terms sum to ¾q. Compute the sum of the odd terms, then multiply by 4/3 to get ζ(2).
Let the series rn = 1/(2n+1)2, as n runs from 0 to infinity. Note that r contains the odd terms of ζ(2).
Consider the function f(x) = x2n. Integrate this from 0 to 1 and obtain 1/(2n+1). Then consider f(x,y) = x2ny2n = (xy)2n. Integrate this over the unit square, as x runs from 0 to 1 and y runs from 0 to 1, and obtain 1/(2n+1)2, or rn. The nth term of the series r has been expressed as a double integral over the unit square.
The series r is the sum of integrals; rewrite this as the integral of the sum. In other words, interchange summation and integration.
With x and y fixed, the sum is (xy)2n, which is a geometric series. The sum is 1 over 1-(xy)2. We need to integrate this over the unit square. Integrate by trig substitution as follows.
x ← sin(u)/cos(v). y ← sin(v)/cos(u).
The denominator, 1-(xy)2, becomes 1 - sin2(u)sin2(v)/cos2(u)cos2(v). This looks like a mess, especially since it's sitting in the denominator, but let's persevere and evaluate the jacobian.
|cos(u) / cos(v)||sin(u)sin(v) / cos(v)2|
|sin(u)sin(v) / cos(u)2,||cos(v) / cos(u)|
Consider the triangle bounded by the u and v axes, and by the line u+v ≤ π/2. The entire hypotenuse, u+v = π/2, maps to the point x = y = 1. The u axis maps to the x axis, the bottom of the unit square, and the v axis maps to the y axis, the left of the unit square. The interior of the triangle becomes the interior of the square. Points near the hypotenuse are squished into the upper right corner, and along the top and right sides of the square. The area of the uv triangle is π2/8. Multiply by 4/3, and ζ(2) = π2/6.
Euler (biography) derived formulas for ζ(2),ζ(4), and others, all without the benefit of modern mathematics. Amazing!