Ta có với mọi số tự nhiên $n$ luôn có:
$\begin{array}{l} \dfrac{1}{{\left( {n + 1} \right)\sqrt n }} = \dfrac{{\sqrt n }}{{\left( {n + 1} \right).n}}\\ = \sqrt n \left( {\dfrac{1}{{\left( {n + 1} \right)n}}} \right)\\ = \sqrt n \left( {\dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right)\\ = \sqrt n \left( {\dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}} \right)\left( {\dfrac{1}{{\sqrt n }} + \dfrac{1}{{\sqrt {n + 1} }}} \right)\\ = \left( {1 + \dfrac{{\sqrt n }}{{\sqrt {n + 1} }}} \right)\left( {\dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}} \right)\\ Do\,\,\dfrac{{\sqrt n }}{{\sqrt {n + 1} }} < 1 \Leftrightarrow 1 + \dfrac{{\sqrt n }}{{\sqrt {n + 1} }} < 2\\ \Rightarrow 1 + \dfrac{{\sqrt n }}{{\sqrt {n + 1} }} < 2\\ \Rightarrow \dfrac{1}{{\left( {n + 1} \right)\sqrt n }} < 2\left( {\dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}} \right) \end{array}$
Áp dụng điều trên ta được:
$\begin{array}{l} \dfrac{1}{{\left( {n + 1} \right)\sqrt n }} < 2\left( {\dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}} \right)\\ \dfrac{1}{{2\sqrt 1 }} < 2\left( {\dfrac{1}{1} - \dfrac{1}{{\sqrt 2 }}} \right)\\ \dfrac{1}{{3\sqrt 2 }} < 2\left( {\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{{\sqrt 3 }}} \right)\\ ....\\ \dfrac{1}{{\left( {n + 1} \right)\sqrt n }} < 2\left( {\dfrac{1}{{\sqrt n }} - \dfrac{1}{{\sqrt {n + 1} }}} \right)\\ \Rightarrow \dfrac{1}{{2\sqrt 1 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\left( {n + 1} \right)\sqrt n }} < 2\left( {1 - \dfrac{1}{{\sqrt {n + 1} }}} \right) < 2\\ \to đpcm \end{array}$
