31st Putnam 1970 Problem B1 Let f(n) = (n 2 + 1)(n 2 + 4)(n 2 + 9) ... (n 2 + (2n) 2 ). Find lim n→∞ f(n) 1/n /n 4 .
Rearranging slightly, we may take g(n) = (1 + (1/n) 2 )(1 + (2/n) 2 )(1 + (3/n) 2 ) ... (1 + (2n/n) 2 ). We have to find lim g(n) 1/n . It is a mistake to approach this algebraically. It is not hard to show that the product of the pair of terms r and 2n+1-r is at least 4. There are n such pairs, so certainly g(n) 1/n > 4, but it is hard to get any further. The key is to take logs. We then see immediately that g(n) 1/n becomes a standard Riemann sum for ∫ 0 2 log(1 + x 2 ) dx. So the limit is simply the integral. x log(1 + x 2 ) differentiates to log(1 + x 2 ) + 2x 2 /(1 + x 2 ). A little reflection then suggests using 2 tan -1 x, which differentiates to 2/(1 + x 2 ). So the complete integral is x log(1 + x 2 ) + 2 tan -1 x - 2x. Evaluating between 0 and 2 gives 2 log 5 + 2 tan -1 2 - 4 = k, say. Then the original limit in the question is e k . That is approximately 4.192. 31st Putnam 1970 © John Scholes jscholes@kalva.demon.co.uk 25 Aug 2001