The sequence of non-negative reals satisfies a_{n+m} ≤ a_{n}a_{m} for all m, n. Show that lim a_{n}^{1/n} exists.

**Solution**

a_{n} ≤ a_{1}a_{n-1} ≤ a_{1}^{2}a_{n-2} ≤ ... ≤ a_{1}^{n}. So a_{n}^{1/n} ≤ a_{1}. All a_{n} are non-negative, so a_{n}^{1/n} ≥ 0. Thus we have established that {a_{n}^{1/n}} is bounded.

Fix n. Take any N > n. Then we may write N = nq + r, with 0 ≤ r < n. We have a_{N} ≤ a_{n}^{q} a_{r} , so a_{N}^{1/N} ≤ a_{n}^{s} a_{r}^{1/N}, where s = q/N = (1 - r/N)/n. But (1 - r/N) tends to 1 as N tends to infinity, as does a_{r}^{1/N}. Hence lim sup a_{N}^{1/N} ≤ a_{n}^{1/n}. This is true for all n, so the sequence cannot have more than one limit point and hence converges.

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002