4th Putnam 1941/B2 solution

4th Putnam 1941

Problem B2

Find:
(1) lim_n→∞ ∑_1≤i≤n 1/√(n² + i²);
(2) lim_n→∞ ∑_1≤i≤n 1/√(n² + i);
(3) lim_n→∞ ∑_1≤i≤n² 1/√(n² + i);

Solution

(1) 1/√(n² + i²) = (1/n) / √(1 + (i/n)²). So the sum is just a Riemann sum for the integral ∫₀¹ dx/√(1 + x²) = sinh^-11 = ln(1 + √2) = 0.8814.

(2) 1/√(n² + i) = (1/n) / √(1 + i/n²). 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/√(n² + i) = n (1/n²) / √(1 + i/n²). Now ∑ (1/n²) / √(1 + i/n²) is just a Riemann sum for ∫₀¹ dx/√(1 + x) = 2√(1 + x) |₀¹ = 2(√2 - 1). So the sum given tends to 2(√2 - 1) n, which diverges to infinity. [Or simpler, there are n² terms, each at least 1/√(2n²) = 1/(n √2 ), so the sum is at least n/√2 which diverges.]

4th Putnam 1941