<s>[INST] Find $\lim_{n \to \infty} \prod_{r=1}^n (n^2+r^2)^{\frac{1}{n}}/n^2$ 
Find $\lim_{n \to \infty} \prod_{r=1}^n \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2}$

Let $L = \lim_{n \to \infty} \prod_{r=1}^n \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2}$
Then $\log(L) = \lim_{n \to \infty} \log(\prod_{r=1}^n \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2})$
$\log(L) = \lim_{n \to \infty} \sum_{r=1}^n \log( \frac{(n^2+r^2)^{\frac{1}{n}}}{n^2})$
$\log(L) = \lim_{n \to \infty} \sum_{r=1}^n(\log( (n^2+r^2)^{\frac{1}{n}})-\log(n^2))$
$\log(L) = \lim_{n \to \infty} \sum_{r=1}^n( \frac{1}{n} \log( n^2(1+(\frac{r}{n})^2)^{})-\log(n^2))$
$\log(L) = \lim_{n \to \infty} \sum_{r=1}^n( \frac{1}{n}( 2\log(n)+\log(1+(\frac{r}{n})^2)^{})-\log(n^2))$
We see that 1 part becomes $\int_{0}^1 \log(1+x^2)dx$
How do I manipulate the remaining terms $\frac{1}{n}2log(n) - log(n^2)$?
 [/INST] Notice that:
$$\prod\nolimits_{r = 1}^{n} \frac {(n^2+r^2)^{\frac{1}{n}}}{n^2}\le \prod\nolimits_{r = 1}^{n}\frac {(n^2+n^2)^{\frac{1}{n}}}{n^2}\le \frac {2n^2}{n^{2n}};$$
hence $L=0$.
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