Push it to the limit

First, note that

$\displaystyle \lim_{n \to\ \infty} \frac{n^2}{\sqrt[n]{(n^2 + 1) \cdot (n^2 + 4) \cdot (n^2 + 9) \cdot \cdot \cdot (n^2 + n^2)}} = \lim_{n \to\ \infty} \sqrt[n]{\frac{n^{2n}}{(n^2 + 1) \cdot (n^2 + 4) \cdot (n^2 + 9) \cdot \cdot \cdot (n^2 + n^2)}}$

Let this limit be $L$ . Splitting up the numerator, we see that

$\displaystyle L = \lim_{n \to\ \infty} \sqrt[n]{\prod_{i = 1}^{n} \frac{n^2}{n^2 + i^2}}$

Taking the natural log of both sides,

$\displaystyle \ln L = \lim_{n \to \infty} \frac 1n \sum_{i = 1}^{n} \ \ln \Bigg(\frac{n^2}{n^2 + i^2} \Bigg) = - \lim_{n \to \infty} \frac 1n \sum_{i = 1}^{n} \ \ln \Bigg(\frac{n^2 + i^2}{n^2} \Bigg) = - \int_{0}^{1} \ln (x^2 + 1) dx$ ,

where in the last step, we used the limit definition of the integral. We evaluate this integral by parts:

Let $u = \ln (x^2 + 1)$ and $dv = 1$ , so this integral is equal to

$\displaystyle x \ln(x^2 + 1) - 2 \int \frac {x^2}{x^2 + 1} dx = x \ln (x^2 + 1) - 2 \int (1 - \frac{1}{x^2 + 1}) dx = x \ln (x^2 + 1) - 2x + 2 \tan ^{-1} x$

Evaluating this from $0$ to $1$ and negating shows that

$\displaystyle \ln L = 2 - \frac{\pi}{2} - \ln 2$ ,

so it follows that $\displaystyle L = {\large{\boxed{\frac 12 e^{2 - \frac {\pi}{2}}}}}$