\limits_{n \to \infty } \left\{ {{{\sqrt n } \over {\sqrt {{n^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 4)}^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 8)}^3}} }} + .... + {{\sqrt n } \over {\sqrt {{{[n + 4(n - 1)]}^3}} }}} \right\}$$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{{\sqrt n } \over {\sqrt {{{(n + 4r)}^3}} }}} $$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{{n\sqrt n } \over {\sqrt {{{(n + 4r)}^3}} }}} $$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{1 \over {\sqrt {{{\left( {1 + {{4r} \over n}} \right)}^{3/2}}} }}} $$

$$ \Rightarrow I = \int_0^1 {{{dx} \over {{{(1 + 4x)}^{3/2}}}}} $$

$$ \Rightarrow I = \left[ {{{{{(1 + 4x)}^{ - 3/2 + 1}}} \over {4( - 3/2 + 1)}}} \right]_0^1$$

$$ \Rightarrow I = {{ - 1} \over 2}[{(1 + 4x)^{ - 1/2}}]_0^1$$

$$ \Rightarrow I = {{ - 1} \over 2}\left[ {{1 \over {\sqrt 5 }} - 1} \right]$$

$$ \Rightarrow I = {{ - 1} \over 2}\left[ {{{1 - \sqrt 5 } \over {\sqrt 5 }}} \right)$$

$$ \Rightarrow I = {{\sqrt 5 - 1} \over {2\sqrt 5 }} = {{\sqrt 5 (\sqrt 5 - 1)} \over {10}} = {{5 - \sqrt 5 } \over {10}}$$19:T51c,Let $$I = \mathop {\lim }\limits_{n \to \infty } \left\{ {{{\sqrt n } \over {\sqrt {{n^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 4)}^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 8)}^3}} }} + .... + {{\sqrt n } \over {\sqrt {{{[n + 4(n - 1)]}^3}} }}} \right\}$$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{{\sqrt n } \over {\sqrt {{{(n + 4r)}^3}} }}} $$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{{n\sqrt n } \over {\sqrt {{{(n + 4r)}^3}} }}} $$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{1 \over {\sqrt {{{\left( {1 + {{4r} \over n}} \right)}^{3/2}}} }}} $$

$$ \Righta

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