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a. Chứng minh rằng, với mọi số nguyên dương k ta đều có
\(\dfrac{1}{{\left( {k + 1} \right)\sqrt k }} < 2\left( {\dfrac{1}{{\sqrt k }} – \dfrac{1}{{\sqrt {k + 1} }}} \right)\)
b. Áp dụng. Chứng minh rằng
\(\dfrac{1}{2} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{4\sqrt 3 }} + … + \dfrac{1}{{\left( {n + 1} \right)\sqrt n }} < 2.\)
:
a. Ta có:
\(\begin{array}{l}\dfrac{1}{{\left( {k + 1} \right)\sqrt k }} = \dfrac{{\sqrt k }}{{\left( {k + 1} \right)k}} = \sqrt k \left( {\dfrac{1}{k} – \dfrac{1}{{k + 1}}} \right)\\ = \sqrt k \left( {\dfrac{1}{{\sqrt k }} + \dfrac{1}{{\sqrt {k + 1} }}} \right)\left( {\dfrac{1}{{\sqrt k }} – \dfrac{1}{{\sqrt {k + 1} }}} \right)\\ = \left( {1 + \dfrac{{\sqrt k }}{{\sqrt {k + 1} }}} \right)\left( {\dfrac{1}{{\sqrt k }} – \dfrac{1}{{\sqrt {k + 1} }}} \right) < 2\left( {\dfrac{1}{{\sqrt k }} – \dfrac{1}{{\sqrt {k + 1} }}} \right)\end{array}\)
b.
\(\begin{array}{l}\dfrac{1}{2} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{4\sqrt 3 }} + … + \dfrac{1}{{\left( {n + 1} \right)\sqrt n }} < 2\left( {1 – \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} – \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{\sqrt 3 }} – \dfrac{1}{{\sqrt 4 }} + … + \dfrac{1}{{\sqrt n }} – \dfrac{1}{{\sqrt {n + 1} }}} \right)\\ = 2\left( {1 – \dfrac{1}{{\sqrt {n + 1} }}} \right) < 2\end{array}\)