Define a_{n} by a_{1} = ln α, a_{2} = ln(α - a_{1}), a_{n+1} = a_{n} + ln(α - a_{n}). Show that lim_{n→∞} a_{n} = α - 1.

**Solution**

Unfortunately, this does not quite work as stated. If α is too small then a_{2} > α and hence a_{3} is undefined. The limit is just under 0.3442. So assume that α is sufficiently large that a_{2} < α.

For all non-zero x we have e^{x} > 1 + x. Hence for all positive x not equal to 1 we have ln x < x - 1.

Suppose that a_{2} = α - 1. Then a_{n} = α - 1 for all n > 2, which gives the result. So assume a_{2} is not equal to α - 1. Then for all n ≥ 2 we have ln(α - a_{n}) < α - a_{n} - 1, so a_{n+1} = a_{n} + ln(α - a_{n}) < α - 1.

Also for n >= 3, we have that α - a_{n} > 1, so ln(α - a_{n}) > 0, so a_{n+1} > a_{n}. Thus for n ≥ 3, a_{n} is a monotonic increasing sequence bounded above by α - 1. So it must converge.

But a_{n+1} - a_{n} = ln(α - a_{n}), so ln(α - a_{n}) tends to 0 and hence a_{n} tends to α - 1.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002