Define an by a0 = 0, an+1 = 1 + sin(an - 1). Find lim (∑0n ai)/n.
Solution
Answer: 1.
Note that 1 - an+1 = sin (1 - an). Put cn = 1 - an. Note that c0 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 = c0 > c1 > c2 > ... > cn > 0. So cn 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 an converges to 1. Hence (∑0n ai)/n converges to 1 also.
© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002