Limit Sum

$\lim_{n \to \infty}\ (\frac {1}{n^2} \sqrt{(n+1)(n-1)} + \frac {1}{n^2} \sqrt{(n+2)(n-2)} +\dots+\frac {1}{n^2} \sqrt{(n+n)(n-n)}) = ?$

This can be expressed as $\lim_{n \to \infty} \displaystyle \sum_{x=1}^n \frac {1}{n^2} \sqrt{(n+x)(n-x)}$

$\lim_{n \to \infty} \displaystyle \sum_{x=1}^n \frac {1}{n^2} \sqrt{n^2-x^2}$

$\lim_{n \to \infty} \displaystyle \sum_{x=1}^n \frac {1}{n} \sqrt{\frac{1}{n^2} (n^2-x^2)}$

$\lim_{n \to \infty} \displaystyle \sum_{x=1}^n \frac {1}{n} \sqrt{1 - \frac{x^2}{n^2}}$

Recall that: $\int^1_0 f(x)\,dx = \lim_{n \to \infty} \displaystyle \sum_{x=1}^n \frac {1}{n} f({\frac {x}{n}})$

Therefore $\lim_{n \to \infty} \displaystyle \sum_{x=1}^n \frac {1}{n} \sqrt{1 - \frac{x^2}{n^2}} = \int^1_0 ∣\sqrt{1-x^2}∣\,dx$

The equation $y = \sqrt{1-x^2}$ can be rearranged to $x^2 + y^2 = 1$

Therefore $\int^1_0 ∣\sqrt{1-x^2}∣\,dx$ is equal to the area of a quarter circle of radius 1.

$\boxed{\frac {\pi}{4}}$