17th Putnam 1957/B2 solution

17th Putnam 1957

Problem B2

The sequence a_n is defined by its initial value a₁, and a_n+1 = a_n(2 - k a_n). For what real a₁ does the sequence converge to 1/k?

Solution

Answer: For a₁ strictly between 0 and 2/k.

Suppose a_n has the opposite sign to k. Then 2 - k a_n is positive, and so a_n+1 also has the opposite sign to k, so we cannot get convergence. Similarly, if a_n = 0, then a_n+1 = 0 and we cannot get convergence. So it is a necessary condition that a₁ should be non-zero and have the same sign as k.

If k is positive and a₁ > 2/k, then a₂ is negative and so the sequence does not converge to 1/k (as above). Similarly, if k is negative and a₁ < 2/k, then a₂ is positive and so we do not have convergence. Thus it is a necessary condition for a₁ to lie strictly between 0 and 2/k.

Suppose a_n = 1/k + h. Then a_n+1 = (1/k + h)(2 - (1 + kh)) = 1/k - kh² (*). This is all we need. But to spell it out, consider first k positive. If a₁ lies strictly between 0 and 1/k, then h lies strictly between -1/k and 1/k. Now (*) shows that for n ≥ 2, a_n ≤ 1/k and is monotonic increasing. Hence it tends to a positive limit. If this limit is L, then referring to the original equation, L = L(2 - kL), so L = 1/k. Similarly for k negative.

17th Putnam 1957