Chứng minh với mọi n ∈ ℤ ta có: \[\frac{1}{{2\sqrt 1 }} + \frac{1}{{3\sqrt 2 }} + \frac{1}{{4\sqrt 3 }} + ... + \frac{1}{{\left( {n + 1} \right)\sqrt n }} < 2\].

Ta có: \(\frac{1}{{\left( {k + 1} \right)\sqrt k }} = \frac{{\sqrt k }}{{k\left( {k + 1} \right)}} = \sqrt k \left( {\frac{1}{k} - \frac{1}{{k + 1}}} \right)\)

        \( = \sqrt k \left( {\frac{1}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }}} \right)\left( {\frac{1}{{\sqrt k }} - \frac{1}{{\sqrt {k + 1} }}} \right)\)

        \( = \left( {1 + \frac{{\sqrt k }}{{\sqrt {k + 1} }}} \right)\left( {\frac{1}{{\sqrt k }} - \frac{1}{{\sqrt {k + 1} }}} \right)\)

Do \(\frac{{\sqrt k }}{{\sqrt {k + 1} }} < 1 \Rightarrow 1 + \frac{{\sqrt k }}{{\sqrt {k + 1} }} < 2 \Rightarrow \frac{1}{{\left( {k + 1} \right)\sqrt k }} < 2\left( {\frac{1}{{\sqrt k }} - \frac{1}{{\sqrt {k + 1} }}} \right)\)

Áp dụng BĐT này, ta có: \(\frac{1}{2} < 2\left( {\frac{1}{1} - \frac{1}{{\sqrt 2 }}} \right)\)

                                         \(\frac{1}{{3\sqrt 2 }} < 2\left( {\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }}} \right)\)

                                         \(\frac{1}{{\left( {n + 1} \right)\sqrt n }} < 2\left( {\frac{1}{{\sqrt n }} - \frac{1}{{\sqrt {n + 1} }}} \right)\)

Cộng tất cả các BĐT trên ta được:

\(\frac{1}{2} + \frac{1}{{3\sqrt 2 }} + \frac{1}{{4\sqrt 3 }} + \ldots + \frac{1}{{\left( {n + 1} \right)\sqrt n }} < 2\left( {\frac{1}{1} - \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 3 }} + \ldots + \frac{1}{{\sqrt n }} - \frac{1}{{\sqrt {n + 1} }}} \right)\).

\(\; \Leftrightarrow \frac{1}{2} + \frac{1}{{3\sqrt 2 }} + \frac{1}{{4\sqrt 3 }} + \ldots + \frac{1}{{\left( {n + 1} \right)\sqrt n }} < 2\left( {1 - \frac{1}{{\sqrt {n + 1} }}} \right) < 2\) (đpcm).