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[hoakute]

Chứng minh rằng với mọi số nguyên dương n-{1} ta luôn có:

$\frac{1}{2}< \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}< \frac{3}{4}$

[pinkyha]

vế đầu:

 

$\dfrac{1}{n+1}+...+\dfrac{1}{n+n} > n\dfrac{1}{n+n}=n.\dfrac{1}{2n}=\dfrac{1}{2}(đpcm)$

Edited by pinkyha, 21-04-2016 - 21:09.