4th Putnam 1941 Problem B2 Find: (1) lim n→∞ ∑ 1≤i≤n 1/√(n 2 + i 2 ); (2) lim n→∞ ∑ 1≤i≤n 1/√(n 2 + i); (3) lim n→∞ ∑ 1≤i≤n 2 1/√(n 2 + i);
(1) 1/√(n 2 + i 2 ) = (1/n) / √(1 + (i/n) 2 ). So the sum is just a Riemann sum for the integral ∫ 0 1 dx/√(1 + x 2 ) = sinh -1 1 = ln(1 + √2) = 0.8814. (2) 1/√(n 2 + i) = (1/n) / √(1 + i/n 2 ). Each term is less than 1/n, so the (finite) sum is less than 1. But each term is at least (1/n) / √(1 + 1/n). So the sum is at least 1/√(1 + 1/n), which tends to 1. Hence the limit of the sum is 1. (3) 1/√(n 2 + i) = n (1/n 2 ) / √(1 + i/n 2 ). Now ∑ (1/n 2 ) / √(1 + i/n 2 ) is just a Riemann sum for ∫ 0 1 dx/√(1 + x) = 2√(1 + x) | 0 1 = 2(√2 - 1). So the sum given tends to 2(√2 - 1) n, which diverges to infinity. [Or simpler, there are n 2 terms, each at least 1/√(2n 2 ) = 1/(n √2 ), so the sum is at least n/√2 which diverges.] 4th Putnam 1941 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002