1+1/4+1/9+… = π²/6 · Fourier coefficients of a sawtooth · π from a circle · runs locally
Basel: Σ 1/n² = π²/6 · Parseval: (1/π)∫₋π^π f(t)² dt = Σ bₙ² · for f(t)=t: LHS=π²/3, RHS=Σ(2/n)² = 4Σ1/n² · so 4Σ1/n²=π²/3 → Σ1/n²=π²/6 · π appears because the sawtooth has period 2π