<s>[INST] Evaluation of $\lim_{n\rightarrow \infty}\frac{1}{n^{2m}}\left[(n^2+1^2)^{m}(n^2+2^2)^m(n^2+3^2)^m..............(2n^2)^{m}\right]^{\frac{1}{n}}$ 
Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n^{2m}}\left[(n^2+1^2)^{m}(n^2+2^2)^m(n^2+3^2)^m..............(2n^2)^{m}\right]^{\frac{1}{n}}$$

$\bf{My\; Try::}$ Let $$L=\lim_{n\rightarrow \infty}\frac{n^{\frac{2m}{n}}}{n^{2m}}\left[\left(\frac{n^2+1^2}{n^2}\right)^m\cdot \left(\frac{n^2+2^2}{n^2}\right)^m\cdot ..........\left(\frac{n^2+n^2}{n^2}\right)^m\right]^{\frac{1}{n}}$$
Altgough I know that we use Riemann Sum of Integral, But I did not understand How can 
I Adjust into $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{r=1}f\left(\frac{r}{n}\right)\cdot \frac{1}{n}$
Help Required, Thanks
 [/INST] If we set
$$ \color{blue}{L}=\lim_{n\to +\infty}\left[\left(1+\frac{1^2}{n^2}\right)\cdot\left(1+\frac{2^2}{n^2}\right)\cdot\ldots\cdot\left(1+\frac{n^2}{n^2}\right)\right]^{1/n}\tag{1}$$
the original limit is just $L^m$. But:
$$ \log L = \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left[1+\left(\frac{k}{n}\right)^2\right]=\int_{0}^{1}\log(1+x^2)\,dx\tag{2}$$
by Riemann sums, hence it is enough to evaluate the last integral. By integration by parts:
$$ \int_{0}^{1}\log(1+x^2)\,dx = \left. x\log(1+x^2)\right|_{0}^{1}-\int_{0}^{1}\frac{2x^2\,dx}{1+x^2}=\log(2)-2+\frac{\pi}{2}, \tag{3}$$
hence:

$$ \color{blue}{L} = \color{red}{2\cdot\exp\left(\frac{\pi}{2}-2\right)}.\tag{4}$$

</s>