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$a=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}< \dfrac{1}{2}$

JakiNatsumi · Accepted Answer

$A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}$

$A=\dfrac{2}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\right)$

$A=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)$

$A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)$

$A=\dfrac{1}{2}.\left(1-\dfrac{1}{2n+1}\right)$

$A=\dfrac{1}{2}-\dfrac{1}{4n+2}< \dfrac{1}{2}\left(dpcm\right)$

Câu trả lời:

$A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}$

$A=\dfrac{2}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\right)$

$A=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)$

$A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)$

$A=\dfrac{1}{2}.\left(1-\dfrac{1}{2n+1}\right)$

$A=\dfrac{1}{2}-\dfrac{1}{4n+2}< \dfrac{1}{2}\left(dpcm\right)$