Define a_{n} by a_{0} = 0, a_{n+1} = 1 + sin(a_{n} - 1). Find lim (∑_{0}^{n} a_{i})/n.

**Solution**

Answer: 1.

Note that 1 - a_{n+1} = sin (1 - a_{n}). Put c_{n} = 1 - a_{n}. Note that c_{0} belongs to the interval (0, 1]. Now for x in (0, 1] we have that sin x < x and sin x is also in (0 , 1]. So it follows (by a trivial induction) that 1 = c_{0} > c_{1} > c_{2} > ... > c_{n} > 0. So c_{n} is a monotonically decreasing sequence bounded below by 0. Hence it must tend to a limit c ≥ 0. But c must satisfy c = sin c. Hence c = 0. Hence a_{n} converges to 1. Hence (∑_{0}^{n} a_{i})/n converges to 1 also.

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002