$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ........ + {n \over {{{(2n + 1)}^2}}}} \right]$$ is equal to :

$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ... + {n \over {{{(2n - 1)}^2}}}} \right]$$ <br><br>$$\mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{{\left( {n + 0} \right)}^2}}} + {n \over {{{\left( {n + 1} \right)}^2}}} + {n \over {{{\left( {n + 2} \right)}^2}}} + ... + {n \over {{{\left( {n + \left( {n - 1} \right)} \right)}^2}}}} \right]$$ <br><br>$$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{n \over {{{(n + r)}^2}}}} = $$$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{n \over {{n^2}{{\left( {1 + {r \over n}} \right)}^2}}}} $$<br><br>$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{1 \over {{{(r/n)}^2} + 2(r/n) + 1}}} $$<br><br>$$ = \int\limits_0^1 {{{dx} \over {{{(x + 1)}^2}}} = \left[ {{{ - 1} \over {(x + 1)}}} \right]_0^1 = {1 \over 2}} $$ <br><br><b>Note :</b> $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{pn} {{1 \over n}f\left( {{r \over n}} \right)} = \int_\alpha ^\beta {f(x)} dx$$<br><br>where, $\alpha = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0$ (as r = 1)<br><br>and $\beta = \mathop {\lim }\limits_{n \to \infty } {r \over n} = p$ (as r = pn) <br><br>Here $\alpha = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0$ (as r = 0) <br><br>and $$\beta = \mathop {\lim }\limits_{n \to \infty } {r \over n} = \mathop {\lim }\limits_{n \to \infty } {{n - 1} \over n}$$ = 1 (as r = n - 1)