17th Putnam 1957 Problem B2 The sequence a n is defined by its initial value a 1 , and a n+1 = a n (2 - k a n ). For what real a 1 does the sequence converge to 1/k? Solution
: For a 1 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 a 1 should be non-zero and have the same sign as k. If k is positive and a 1 > 2/k, then a 2 is negative and so the sequence does not converge to 1/k (as above). Similarly, if k is negative and a 1 < 2/k, then a 2 is positive and so we do not have convergence. Thus it is a necessary condition for a 1 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 - kh 2 (*). This is all we need. But to spell it out, consider first k positive. If a 1 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 © John Scholes jscholes@kalva.demon.co.uk 25 Feb 2002