62nd Putnam 2001 Problem B3 Let @n denote the closest integer to √n. Find ∑ (2 @n + 2 -@n )/2 n , where the sum is taken from n = 1 to n = infinity. Solution
: 3. We need to find a more useful expression for @n. We have (n + 1/2) 2 = n 2 + n + 1/4, which lies between the integers n(n+1) and n(n+1)+1. Thus the 2n values @m for m = (n-1)n+1, (n-1)n+2, ... , (n- 1)n+2n=n(n+1) are all n. The sum is obviously absolutely convergent, so we can consider ∑ 2 @n /2 n and ∑ 2 -@n /2 n separately. The first is (2 0 + 2 -1 ) + (2 -1 + 2 -2 + 2 -3 + 2 -4 ) + (2 -4 + ... + 2 -9 ) + ... = (1 + 1/2 + 1/4 + ... ) + (1/2 + 1/2 4 + 1/2 9 + 1/2 16 + ... ). The second sum is (1/2 2 + 1/2 3 ) + (1/2 5 + ... + 1/2 8 ) + (1/2 10 + ... + 1/2 15 ) + ... . Now (1 + 1/2 + 1/4 + ... ) = 2, and the other two brackets together give (1/2 + 1/4 + 1/8 + ... ) = 1. 62nd Putnam 2001 © John Scholes jscholes@kalva.demon.co.uk 16 Dec 2001