The sequences a_{n}, b_{n}, c_{n} of positive reals satisfy: (1) a_{1} + b_{1} + c_{1} = 1; (2) a_{n+1} = a_{n}^{2} + 2b_{n}c_{n}, b_{n+1} = b_{n}^{2} + 2c_{n}a_{n}, c_{n+1} = c_{n}^{2} + 2a_{n}b_{n}. Show that each of the sequences converges and find their limits.

**Solution**

a_{n+1} + b_{n+1} + c_{n+1} = (a_{n} + b_{n} + c_{n})^{2}, so by a trivial induction a_{n} + b_{n} + c_{n} = 1. There appears to be symmetry between the three sequences, so we conjecture that each converges to 1/3.

Suppose a_{n} ≤ b_{n} ≤ c_{n}. We have a_{n+1} = a_{n}^{2} + 2b_{n}c_{n} ≤ a_{n}c_{n} + b_{n}c_{n} + c_{n}c_{n} = c_{n}(a_{n} + b_{n} + c_{n}) = c_{n}. Similarly, b_{n+1} = b_{n}_{2} + 2a_{n}c_{n} ≤ b_{n}c_{n} + a_{n}c_{n} + c_{n}^{2} = c^{n}, and c_{n+1} = c_{n}^{2} + 2a_{n}b_{n} ≤ c_{n}^{2} + a_{n}c_{n} + b_{n}c_{n} = c_{n}. Hence the largest of a_{n+1}, b_{n+1}, c_{n+1} is no bigger than the largest of a_{n}, b_{n}, c_{n}. 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 b_{n} - a_{n} = h ≥ 0, c_{n} - b_{n} = k ≥ 0. Then a_{n+1} - b_{n+1} = (a_{n} - b_{n})(a_{n} + b_{n} - 2c_{n}), so |a_{n+1} - b_{n+1}| = h(h + 2k) ≤ (h + k)^{2}. Similarly, |b_{n+1} - c_{n+1}| = |b_{n} - c_{n}| |b_{n} + c_{n} - 2a_{n}| = k(2h + k) ≤ (h + k)^{2}, and |c_{n+1} - a_{n+1}| = |c_{n} - a_{n}| |c_{n} + a_{n} - 2b_{n}| = |(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 a_{1}, b_{1}, c_{1} are all positive and hence, by a trivial induction, a_{n}, b_{n}, c_{n} 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 a_{n}, b_{n}, c_{n} all tend to 1/3.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002