Here's a thought: Archimedes's approximation approaches the target. That is, the perimeter of each subsequent approximation is closer to that of the circle than

Here's a thought: Archimedes's approximation approaches the target. That is, the perimeter of each subsequent approximation is closer to that of the circle than all that came before. The side of a regular polygon is √(sin²(τ/n)+(1−cos(τ/n))²)=√(2(1−cos(τ/n))), while the corresponding arc is τ/n. d/dn τ/n−√(2(1−cos(τ/n))) = −(τ/n²+sin(τ/n)/√(2(1−cos(τ/n)))) < 0, so the difference is strictly decreasing (note that sin(τ/n)>0 when n>2). So there's your lower bound for the difference. You'd still have to prove the limit is at 0, though.