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

Separating even indices $n=2m$ from odd indices $n=2k+1$ turns the even part into $\tfrac14\zeta(2)$ itself; solving for $\zeta(2)$ reduces the whole problem to the sum over odd squares only, which matches a natural double integral in the next step.

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