17th Putnam 1957/A6 solution

17th Putnam 1957

Problem A6

Define a_n by a₁ = ln α, a₂ = ln(α - a₁), 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 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 e^x > 1 + x. Hence for all positive x not equal to 1 we have ln x < x - 1.

Suppose that a₂ = α - 1. Then a_n = 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 - 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.

17th Putnam 1957