Tính \(\lim \left( {\frac{1}{{2\sqrt 1  + 1\sqrt 2 }} + \frac{1}{{3\sqrt 2  + 2\sqrt 3 }} + ... + \frac{1}{{\left( {n + 1} \right)\sqrt n  + n\sqrt {n + 1} }}} \right)\)

\(\lim \left( {\frac{1}{{2\sqrt 1  + 1\sqrt 2 }} + \frac{1}{{3\sqrt 2  + 2\sqrt 3 }} + ... + \frac{1}{{\left( {n + 1} \right)\sqrt n  + n\sqrt {n + 1} }}} \right)\)

\( = \lim \left( {\frac{{2\sqrt 1  - 1\sqrt 2 }}{{2 \cdot 1}} + \frac{{3\sqrt 2  - 2\sqrt 3 }}{{3 \cdot 2}} + ... + \frac{{\left( {n + 1} \right)\sqrt n  - n\sqrt {n + 1} }}{{\left( {n + 1} \right) \cdot n}}} \right)\)

\( = \lim \left( {1 - \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }} + ... + \frac{1}{{\sqrt n }} - \frac{1}{{\sqrt {n + 1} }}} \right)\)

\( = \lim \left( {1 - \frac{1}{{\sqrt {n + 1} }}} \right) = 1\).

Trả lời: 1.