10th Putnam 1950 Problem A3 The sequence a n is defined by a 0 = α, a 1 = β, a n+1 = a n + (a n-1 - a n )/(2n). Find lim a n .
a n+1 - a n = -1/2n (a n - a n-1 ), hence a n+1 - a n = (-1) n /(2·4·6 ... .2n) (a 1 - a 0 ). Then a n = (a n - a n-1 ) + (a n-1 - a n-2 ) + ... + (a 1 - a 0 ) + a 0 . Hence the limit is β + (β - α)/2 - (β - α)/2·4 + (β - α)/2·4·6 - ... . But 1/2 -1/2·4 + 1/2·4·6 - ... = 1 - 1/√e, so the limit is α + (β - α)/√e. 10th Putnam 1950 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002