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}.

**Solution**

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.

© John Scholes

jscholes@kalva.demon.co.uk

25 Aug 2001