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≤n2} 1/√(n^{2} + i);

**Solution**

(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.]

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002