The Basel Problem: Lighthouses Sum to π²/6

1 + 1/4 + 1/9 + ⋯ = π²/6 ≈ 1.6449 — and the reason π appears is circles.
Lighthouses on the Number Line
Brightness 1/n² per lighthouse
Number of lighthouses
1
1.0000
Σ 1/n² (n = 1 to 1)
π²/6 ≈ 1.6449
60.8% of limit
Σ 1/n² → π²/6 ≈ 1.6449
At n = 100: sum ≈ 1.6350 (99.4% of limit)
At n = 200: sum ≈ 1.6399 (99.7% of limit)
Converges slowly — error ≈ 1/n

Convergence curve — partial sums vs. π²/6

Why does π appear?
The lighthouse proof (Beukers, Calabi, Kolk 1993): Place lighthouses at 1, 2, 3, … on a line. Total apparent brightness = Σ 1/n². Now imagine the same lighthouses on a circle — circles carry π. As you expand circles and flatten them back into the number line, the two-sided sum 2Σ 1/n² equals π²/3, so one-sided Σ 1/n² = π²/6. That's why π appears: because circles carry π, and circles are the natural home of these sums.