[toán 11] Quy nạp toán hoc.

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lethithuydungdn

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N

noinhobinhyen

$1 > \dfrac{1}{\sqrt{n}}$

$ \dfrac{1}{\sqrt{2}} > \dfrac{1}{\sqrt{n}}$

$ \dfrac{1}{\sqrt{3}} > \dfrac{1}{\sqrt{n}}$

...

$\Rightarrow 1+ \dfrac{1}{\sqrt{2}}+ \dfrac{1}{\sqrt{3}}+...+ \dfrac{1}{\sqrt{n}} >
n \dfrac{1}{\sqrt{n}} = \sqrt{n}$
 
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