Question

A4
Să se arate că pentru neN*:
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Answer 1

$\it \dfrac{1}{1}+\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt3}+\ ...\ +\dfrac{1}{\sqrt n}\geq\underbrace{\dfrac{1}{\sqrt n}+\dfrac{1}{\sqrt n}+\dfrac{1}{\sqrt n}+\ ...\ \dfrac{1}{\sqrt n}}_{n\ termeni}=\dfrac{n}{\sqrt n}=\\ \\ \\ =\dfrac{\sqrt n\cdot\sqrt n}{\sqrt n}=\sqrt n$