The sequences an, bn, cn of positive reals satisfy: (1) a1 + b1 + c1 = 1; (2) an+1 = an2 + 2bncn, bn+1 = bn2 + 2cnan, cn+1 = cn2 + 2anbn. Show that each of the sequences converges and find their limits.
Solution
an+1 + bn+1 + cn+1 = (an + bn + cn)2, so by a trivial induction an + bn + cn = 1. There appears to be symmetry between the three sequences, so we conjecture that each converges to 1/3.
Suppose an ≤ bn ≤ cn. We have an+1 = an2 + 2bncn ≤ ancn + bncn + cncn = cn(an + bn + cn) = cn. Similarly, bn+1 = bn2 + 2ancn ≤ bncn + ancn + cn2 = cn, and cn+1 = cn2 + 2anbn ≤ cn2 + ancn + bncn = cn. Hence the largest of an+1, bn+1, cn+1 is no bigger than the largest of an, bn, cn. An exactly similar argument works for the smallest. Hence the largest forms a monotonic decreasing sequence which is bounded below and the smallest forms a monotonic increasing sequence which is bounded above.
Let bn - an = h ≥ 0, cn - bn = k ≥ 0. Then an+1 - bn+1 = (an - bn)(an + bn - 2cn), so |an+1 - bn+1| = h(h + 2k) ≤ (h + k)2. Similarly, |bn+1 - cn+1| = |bn - cn| |bn + cn - 2an| = k(2h + k) ≤ (h + k)2, and |cn+1 - an+1| = |cn - an| |cn + an - 2bn| = |(h + k)(k - h)| ≤ (h + k)2. So for n+1 the difference between the biggest and the smallest is the square of the difference for n. But a1, b1, c1 are all positive and hence, by a trivial induction, an, bn, cn are positive. Their sum is 1 so the difference between the biggest and smallest must be less than 1. Hence the difference tends to zero. Hence an, bn, cn all tend to 1/3.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002