Let f(n) = (n2 + 1)(n2 + 4)(n2 + 9) ... (n2 + (2n)2). Find limn→∞ f(n)1/n/n4.
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 ∫02 log(1 + x2) dx. So the limit is simply the integral.
x log(1 + x2) differentiates to log(1 + x2) + 2x2/(1 + x2). A little reflection then suggests using 2 tan-1x, which differentiates to 2/(1 + x2). So the complete integral is x log(1 + x2) + 2 tan-1x - 2x. Evaluating between 0 and 2 gives 2 log 5 + 2 tan-12 - 4 = k, say. Then the original limit in the question is ek. That is approximately 4.192.
31st Putnam 1970
© John Scholes
25 Aug 2001