Mathematical Reflections 2015, Issue 3 - Problem U340

Problem:
Define $$A_n=\dfrac{n}{n^2+1^2}+\dfrac{n}{n^2+2^2}+\ldots+\dfrac{n}{n^2+n^2}.$$
Find $$\lim_{n \to \infty} n\left(n\left(\dfrac{\pi}{4}-A_n\right)-\dfrac{1}{4}\right).$$

Proposed by Yong Xi Wang, China

Solution:
We have $$nA_n=B_n,$$ where $$B_n=\dfrac{1}{1+\left(\frac{1}{n}\right)^2}+\ldots+\dfrac{1}{1+\left(\frac{n}{n}\right)^2}.$$
$B_n$ is the Riemann lower sum of $f(x)=\dfrac{1}{1+x^2}$ over $[0,1]$ with partition $$P=\left\{\left[0,\dfrac{1}{n}\right],\left[\dfrac{1}{n},\dfrac{2}{n}\right],\ldots,\left[\dfrac{n-1}{n},1\right]\right\}.$$
Therefore, $$\lim_{n \to \infty} B_n=\int_{0}^1 \dfrac{1}{1+x^2} dx=\left[\arctan x\right]_{0}^1=\dfrac{\pi}{4}.$$
It follows that $$\lim_{n \to \infty} n\left(n\left(\dfrac{\pi}{4}-A_n\right)-\dfrac{1}{4}\right)=\lim_{n \to \infty} n \cdot \lim_{n \to \infty} \left(\dfrac{n\pi}{4}-nA_n-\dfrac{1}{4}\right)=\infty.$$


Note. The result obtained is different from the official solution. See the official solution.