Define an by a1 = ln α, a2 = ln(α - a1), an+1 = an + ln(α - an). Show that limn→∞ an = α - 1.
Unfortunately, this does not quite work as stated. If α is too small then a2 > α and hence a3 is undefined. The limit is just under 0.3442. So assume that α is sufficiently large that a2 < α.
For all non-zero x we have ex > 1 + x. Hence for all positive x not equal to 1 we have ln x < x - 1.
Suppose that a2 = α - 1. Then an = α - 1 for all n > 2, which gives the result. So assume a2 is not equal to α - 1. Then for all n ≥ 2 we have ln(α - an) < α - an - 1, so an+1 = an + ln(α - an) < α - 1.
Also for n >= 3, we have that α - an > 1, so ln(α - an) > 0, so an+1 > an. Thus for n ≥ 3, an is a monotonic increasing sequence bounded above by α - 1. So it must converge.
But an+1 - an = ln(α - an), so ln(α - an) tends to 0 and hence an tends to α - 1.
17th Putnam 1957
© John Scholes
5 Mar 2002