[ _n [ 1 n + n (n + 1)^2 + n (n + 2)^2 + ........

Question

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

Accepted Answer

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

$$\mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{{\left( {n + 0} \right)}^2}}} + {n \over {{{\left( {n + 1} \right)}^2}}} + {n \over {{{\left( {n + 2} \right)}^2}}} + ... + {n \over {{{\left( {n + \left( {n - 1} \right)} \right)}^2}}}} \right]$$

$$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{n \over {{{(n + r)}^2}}}}  = $$ $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{n \over {{n^2}{{\left( {1 + {r \over n}} \right)}^2}}}} $$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{1 \over {{{(r/n)}^2} + 2(r/n) + 1}}} $$ = \int\limits_0^1 {{{dx} \over {{{(x + 1)}^2}}} = \left[ {{{ - 1} \over {(x + 1)}}} \right]_0^1 = {1 \over 2}} $Note :$\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$where,$\alpha  = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0$(as r = 1)

and$\beta  = \mathop {\lim }\limits_{n \to \infty } {r \over n} = p$(as r = pn)

Here$\alpha  = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0$(as r = 0)

and$\beta  = \mathop {\lim }\limits_{n \to \infty } {r \over n} = \mathop {\lim }\limits_{n \to \infty } {{n - 1} \over n}$

Solutions · 1