Tính:
$$\lim_{n \to \infty } \frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+...+\frac{n}{n^2+n^2}$$
Ta có: $$\mathop {\lim }\limits_{n \to \infty } \frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + ... + \frac{n}{{{n^2} + {n^2}}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{\left( {\frac{k}{n}} \right)}^2}}}} $$
Đây chính là tổng tích phân của hàm số $f\left( x \right) = \frac{1}{{1 + {x^2}}},x \in \left( {0;1} \right)$
Do đó: $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}}}} = \int\limits_0^1 {\frac{1}{{1 + {x^2}}}dx = \left. {arctgx} \right|} _0^1 = \frac{\pi }{4}$$
