Official solution
\[\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{n \over {{n^2} + {r^2}}}} \]
\[\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{1 \over {n\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}} = \int\limits_0^2 {{{dx} \over {1 + {x^2}}}} } = {\tan ^{ - 1}}2\]