Đáp án:
57 C
Giải thích các bước giải:
57.
Ta có: \(\dfrac{1}{{(n + 1)\sqrt n + n\sqrt {n + 1} }} \) \(=\dfrac{1}{{\sqrt {n\left( {n + 1} \right)} \left( {\sqrt {n + 1} + \sqrt n } \right)}}\) \( = \dfrac{{\sqrt {n + 1} - \sqrt n }}{{\sqrt {n\left( {n + 1} \right)} \left( {\sqrt {n + 1} + \sqrt n } \right)\left( {\sqrt {n + 1} - \sqrt n } \right)}}\) \( = \dfrac{{\sqrt {n + 1} - \sqrt n }}{{\sqrt {n\left( {n + 1} \right)} \left( {n + 1 - n} \right)}}\) \( = \dfrac{{\sqrt {n + 1} - \sqrt n }}{{\sqrt n .\sqrt {n + 1} }} = \dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}\)
\( \Rightarrow {u_n} = \dfrac{1}{{2\sqrt 1 + \sqrt 2 }} + \dfrac{1}{{3\sqrt 2 + 2\sqrt 3 }} + ... + \dfrac{1}{{(n + 1)\sqrt n + n\sqrt {n + 1} }} \) \(= \dfrac{1}{{\sqrt 1 }} - \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 3 }} + ... + \dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}\)
Suy ra \({u_n} = 1 - \dfrac{1}{{\sqrt {n + 1} }} \Rightarrow \lim {u_n} = \lim \left( {1 - \dfrac{1}{{\sqrt {n + 1} }}} \right) = 1\) do \(\lim \dfrac{1}{{\sqrt {n + 1} }} = 0\)
