: 4 + π. We claim that: (1) the sequence diverges for |y| > 1; (2) the sequence diverges for |y|  1 + √(1-y 2 ); (3) the sequence converges for |y|  1, then a n+1 - a n > (y 2 - 1)/2 > 0, which establishes (1). In case (2), take ε > 0, such that |x| > 1 + √(1 - y 2 ) + ε. Then (x - 1) 2 > 1 - y 2 + ε 2 , so a 1 - x > ε 2 /2 (using (*) ). Hence a 1 > x > 1 + √(1 - y 2 ) + ε. So by a simple induction a n+1 - a n > ε 2 /2 for all n, which establishes (2). In case (3), assume first that x is positive. x < 1 + √(1-y 2 ) implies that (x - 1) 2 < 1 - y 2 , so by (*) a 1 < x. So by a simple induction a n+1 < a n for all x. But all a n are positive, so convergence follows. In the case x negative the sequence is the same for all later terms as that started by -x, so we have convergence in this case also. 53rd Putnam 1992 © John Scholes jscholes@kalva.demon.co.uk 7 Jan 2001
53rd Putnam 1992 Problem B3 Let S be the set of points (x, y) in the plane such that the sequence a n defined by a 0 = x, a n+1 = (a n 2 + y 2 )/2 converges. What is the area of S? Solution