Worked solution: The Beukers–Calabi–Kolk double-integral proof (1993)

Step 1 of 6: Split into odd and even squares

In plain words

Sort the terms of $\zeta(2)$ into two piles by whether the index is odd or even, the way you might sort socks by color. The even pile turns out to be an exact quarter-sized copy of the whole sum, so subtracting it away and solving for $\zeta(2)$ leaves only the odd pile — the sum over $1,3,5,\dots$ — to understand.

\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}+\frac14\sum_{m=1}^{\infty}\frac{1}{m^2}\ \Longrightarrow\ \zeta(2)=\frac43\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}

Detailed analysis

Write $\zeta(2)=\sum_{n\ \mathrm{odd}}\frac{1}{n^2}+\sum_{n\ \mathrm{even}}\frac{1}{n^2}$ . Setting $n=2m$ in the even part gives $\sum_{m=1}^{\infty}\frac{1}{(2m)^2}=\frac{1}{4}\sum_{m=1}^{\infty}\frac{1}{m^2}=\frac{1}{4}\zeta(2)$ : the even-indexed tail is a scaled copy of the very series being evaluated. Absorbing it yields $\zeta(2)-\frac{1}{4}\zeta(2)=\frac{3}{4}\zeta(2)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$ , hence $\zeta(2)=\frac{4}{3}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$ .

This parity split, used in the Beukers–Calabi–Kolk paper and in its predecessors, is what makes the rest of the proof possible: the denominator $1-x^2y^2$ appearing in Step 2 expands into a geometric series in $(xy)^2$ , whose powers are exactly the even monomials $x^{2k}y^{2k}$ needed to produce the odd denominators $(2k+1)^2$ after integration. With all squares one would instead need the denominator $1-xy$ , which leads to a genuinely harder singular integral.

The split is an algebraic identity, not an approximation, and it is reversible — knowing the odd-square sum is equivalent to knowing $\zeta(2)$ itself.

Terms in this step

$\zeta(2)$: Standard shorthand for the value at $2$ of the Riemann zeta function, $\zeta(s)=\sum_{n=1}^\infty 1/n^s$ — here the infinite sum $\sum 1/n^2$ from the Basel problem.

Knowledge used in this step

Numerical series and convergence tests