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(D) The given expression is $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^{2n} {\frac{n}{{{n^2} + {r^2}}}}$.We can rewrite the general t

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$	an^{-1}(2)$

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(D) The given expression is $\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r = 1}^{2n} {\frac{n}{{{n^2} + {r^2}}}}$.We can rewrite the general term as $\frac{n}{n^2(1 + (r/n)^2)} = \frac{1}{n} \cdot \frac{1}{1 + (r/n)^2}$.Using the definition of the definite integral as the limit of a sum,$\mathop {\lim }\limits_{n 	o \infty } \sum\limits_{r=1}^{kn} \frac{1}{n} f(\frac{r}{n}) = \int_0^k f(x) dx$.Here,$f(x) = \frac{1}{1+x^2}$ and the upper limit is $2$ (since $r$ goes up to $2n$).Thus,the integral becomes $\int_0^2 \frac{1}{1+x^2} dx$.Evaluating the integral,we get $[	an^{-1}(x)]_0^2 = 	an^{-1}(2) - 	an^{-1}(0) = 	an^{-1}(2)$.

$\mathop {\lim }\limits_{n \to \infty } \,\left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \frac{n}{{{n^2} + {3^2}}} + ... + \frac{n}{{{n^2} + {{(2n)}^2}}}} \right)$ is equal to

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