John Derbyshire has a little maths problem at the tail of his August Diary in National Review Online. Here it is:

Draw a circle of radius 1 unit (foot, meter, mile, whatever). Inscribe an equilateral triangle in the circle, the three corners all on the circle's circumference. Now inscribe a smaller circle in the triangle, its circumference just touching the sides of the triangle at their mid-points.

Starting with that smaller inner circle, repeat the process; but this time inscribe a square instead of a triangle. Then inscribe a yet smaller circle in the square, its circumference just touching the sides of the square at their mid-points.

Starting with this yet smaller circle, inscribe a regular pentagon in it ... and then a circle inside the pentagon, touching its sides. Keep going in this fashion, with a hexagon, a heptagon, and so on.

The circle you inscribe inside an N-gon will be the (N-1)-th circle you've drawn. (The circle inscribed in the square, for example, is the third one you've drawn.) The radii of all the circles you have drawn up to that point will form a sequence of (N-1) numbers, each smaller than the one before: 1, r₂, r₃, r₄, ... r_N-1. What is the limit of this sequence?

First, let's make sure it has a limit. The ratio of the radii of the incircle to the circumcircle (call 'em r and R respectively) of a regular n-gon is r/R = cos(pi/n). So the second circle (the one inside the triangle) is cos(pi/3) = one half the radius of the circumscribing circle. The next one has cos(pi/4) = 1/Sqrt[2] the radius of that. The next one is cos(pi/5) = (1 + Sqrt[5])/4 of that and so on. So the radius of the nth circle is the product of cos(pi/k) as k goes from 3 to n. Does this converge in the limit? Yes. Why?

Recast the product. Take the natural log of each term and switch the product to a sum. If you take e, the base of the natural logarithms, to this power then you get the same answer as the product. Does the series converge? Use d'Alembert's ratio test. In the limit as k goes to infinity, is the ratio of the k+1th term to the kth term less than unity? Yup. Series converges. In the interval in which we're interested, log of cos(x) as x goes to zero is a monotonically increasing, strictly negative function so d'Alembert is applicable and gives the answer that the sum converges.

There's no closed-form solution for the radius of the inner circle in the limit. Its numerical value is approximately 0.1149420448532, but using the product of cosines formula converges horribly slowly. However, using some evil jiggery-pokery and the product of terms -> sum of logs technique adumbrated above, you can get a fast converging series using, mirabile dictu, the Riemann zeta function.

UPDATE: What the hell was I thinking? The ratio test is useless. Limit as k^th/k+1^th term goes to infinity is unity, and d'Alembert has nothing to say in this case. Fortunately, the Integral Test rides to the rescue. Recast the sum as minus sum of log of sec of pi/x for x = 3 to infinity, check that in the limit log of sec of 1/x goes to zero (it does), and check that in the region we're talking about, the integral is bounded. It is. There's no closed form solution for the integral of the log of an elementary trig function, but numerically it converges. So we're OK.