CM : $(1-\dfrac{1}{2^2})(1-\dfrac{1}{3^2}.....(1-\dfrac{1}{n^2}$ $N \geq 2 $ và là số tự nhiện

C

congchuaanhsang

$VT=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}........\dfrac{n^2-1}{n^2}$

=$\dfrac{n+1}{2n}$ > $\dfrac{1}{2}$
 
R

ronaldover7

1-$\frac{1}{2^2}$=$\frac{2^2-1}{2^2}$=$\frac{(2-1)(2+1)}{2^2}$=$\frac{1.3}{2^2}$
1-$\frac{1}{3^2}$=$\frac{3^2-1}{3^2}$=$\frac{(3-1)(3+1)}{3^2}$=$\frac{2.4}{3^2}$
................
1-$\frac{1}{n^2}$=$\frac{n^2-1}{n^2}$=$\frac{(n-1)(n+1)}{n^2}$
\Rightarrow (1-$\frac{1}{2^2}$)(1-$\frac{1}{3^2}$)....(1-$\frac{1}{n^2}$)
=$\frac{1.3}{2^2}$$\frac{2.4}{3^2}$...........$\frac{(n-1)(n+1)}{n^2}$
=$\frac{n+1}{2n}$ > $\frac{1}{2}$​
 
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