A delicious limit

If $X_n \; = \; \frac{1}{n^4}\prod_{k=1}^{2n} (n^2+k^2)^{\frac1n} \; = \; \prod_{k=1}^{2n}\left(1 + \frac{k^2}{n^2}\right)^{\frac1n}$ then $\ln X_n \; =\; \frac{1}{n}\sum_{k=1}^{2n}\ln\left(1 + \frac{k^2}{n^2}\right)$ and hence $\lim_{n \to \infty} \ln X_n \; = \; \int_0^2 \ln(1 + x^2)\,dx \; = \; \Big[x\ln(1+x^2) - 2x + 2\tan^{-1}x\Big]_0^2 \; = \; 2\ln5 - 4 + 2\tan^{-1}2$ and hence $\lim_{n \to \infty} X_n \; = \; 5^2 e^{2\tan^{-1}2 - 2^2}$ making the answer $2+5 = \boxed{7}$ .