The value of [ _n 1 n_r = 0^2n - 1 n^2 n^2 + 4r^2 ] is :

Question

The value of

$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}} $$ is :

Accepted Answer

$$L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{1 \over {1 + 4{{\left( {{r \over n}} \right)}^2}}}} $$

$$ \Rightarrow L = \int\limits_0^2 {{1 \over {1 + 4{x^2}}}dx} $$

$$ \Rightarrow L = \left. {{1 \over 2}{{\tan }^{ - 1}}(2x)} \right|_0^2 \Rightarrow L = {1 \over 2}{\tan ^{ - 1}}4$$

Solutions · 1