The sequence an is defined by its initial value a1, and an+1 = an(2 - k an). For what real a1 does the sequence converge to 1/k?
Answer: For a1 strictly between 0 and 2/k.
Suppose an has the opposite sign to k. Then 2 - k an is positive, and so an+1 also has the opposite sign to k, so we cannot get convergence. Similarly, if an = 0, then an+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 an = 1/k + h. Then an+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, an ≤ 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
© John Scholes
25 Feb 2002