Question

2/5 < 1/1×3+1/3×5+1/5×7+.....+1/2013×2015 <1/2 urgent! Ms

Answer 1

$\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{2013\cdot2014}=\dfrac12\cdot\left(\dfrac11-\dfrac13+\dfrac13-\dfrac15+\dfrac15-\dfrac17+...$

$...+\dfrac{1}{2013}-\dfrac{1}{2015})\right)=\dfrac12\cdot\dfrac{2014}{2015}=\dfrac{1012}{2015}$

Inegalitatea din enunt devine acum

$\dfrac25<\dfrac{1012}{2015}<\dfrac12$

care adusa la acelasi numarator devine:

$\dfrac{1012}{2530}<\dfrac{1012}{2015}<\dfrac{1012}{2014}$ care este adevarata (dintre doua fractii cu acelasi numarator este mai mare cea cu numitorul mai mic.)