Σ\(k^{2}\)=n(n+1)(2n+1)/6だから、
(1/\(n^{2}\))Σ(\(n^{2}\)-\(k^{2}\))
=(1/\(n^{2}\))(\(n^{3}\)-(2\(n^{3}\)+3\(n^{2}\)+n)/6)
=(1/\(n^{2}\))(4\(n^{3}\)-3\(n^{2}\)-n)/6
=(4n-3-\(\frac{1}{n}\))/6->∞ (as n->∞)