You have something resembling a Riemann sum. Multiply through the summand by 1/n², then you can write
[tex]\displaystyle \lim_{n\to\infty} \frac1n \sum_{r=0}^{2n-1} \frac{1}{1+4\left(\frac rn\right)^2} = \int_0^2 \frac{dx}{1+4x^2} = \boxed{\frac12 \tan^{-1}(4)}[/tex]
