Evaluate lim_{n→∞} 1/n^{5} ∑ (5 r^{4} - 18 r^{2} s^{2} + 5 s^{4}), where the sum is over all r, s satisfying 0 < r, s ≤ n.

**Solution**

Answer: - 1.

This seems curiously difficult for a qu 1, which makes me think I am missing something!

We need: ∑_{1}^{n} r^{2} = 1/3 n^{3} + 1/2 n^{2} + O(n), and ∑_{1}^{n} r^{4} = 1/5 n^{5} + 1/2 n^{4} + O(n^{3}).

Applying these we get that the sum given is 5 (1/5 n^{5} + 1/2 n^{4} + O(n^{3}) ) n - 18 (1/3 n^{3} + 1/2 n^{2} + O(n) )^{2} + 5 (1/5 n^{5} + 1/2 n^{4} + O(n^{3}) ) n = (n^{6} + 5/2 n^{5} + O(n^{4}) ) - (2n^{6} + 6n^{5} + O(n^{4}) ) + (n^{6} + 5/2 n^{5} + O(n^{4}) ) = - n^{5} + O(n^{4}). So the required limit is - 1.

The tricky part is proving the relation for ∑ r^{4}. A fairly trivial induction gives that ∑ r(r+1)(r+2)...(r+s) = 1/(s+2) r(r+1)...(r+s+1). Hence ∑ r^{k} = 1/(k+1) n^{k+1} + O(n^{k}). Also ∑(r^{4} + 6r^{3} + O(r^{2}) ) = 1/5 (n^{5} + 10 n^{4} +O(n^{3}) ). Hence ∑ r^{4} = 1/5 n^{5} + 2n^{4} + O(n^{3}) - 6(1/4 n^{4} + O(n^{3}) ) - ∑ O(n^{2}) = 1/5 n^{5} + 1/2 n^{4} + O(n^{3}). The relation for ∑ r^{2} can be obtained similarly, although most people probably remember the exact formula 1/6 n(n+1)(2n+1), which is easy to prove by induction.

© John Scholes

jscholes@kalva.demon.co.uk

16 Jan 2001