a) $(1+\frac{1}{1.3})(1+\frac{1}{2.4})...(1+\frac{1}{n(n+2)})<2 $ Chú ý $1+\frac{1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$. Do đó $(1+\frac{1}{1.3})(1+\frac{1}{2.4})...(1+\frac{1}{n(n+2)})=\frac{2^2}{1.3}.\frac{3^2}{2.4}\cdots\frac{(n+1)^2}{n(n+2)}=\frac{2.(n+1)!^2}{n!.(n+2)!}=\frac{2(n+1)}{n+2}<2 $
a) $(1+\frac{1}{1.3})(1+\frac{1}{2.4})...(1+\frac{1}{n(n+2)})<2 $ Chú ý $1+\frac{1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$. Do đó $(1+\frac{1}{1.3})(1+\frac{1}{2.4})...(1+\frac{1}{n(n+2)})=\frac{2^2}{1.3}.\frac{3^2}{2.4}\cdots\frac{(n+1)^2}{n(n+2)}=\frac{2.(n+1)!^2}{n!.(n+2)!}=\frac{2}{n+2}<2 $
a) $(1+\frac{1}{1.3})(1+\frac{1}{2.4})...(1+\frac{1}{n(n+2)})<2 $ Chú ý $1+\frac{1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$. Do đó $(1+\frac{1}{1.3})(1+\frac{1}{2.4})...(1+\frac{1}{n(n+2)})=\frac{2^2}{1.3}.\frac{3^2}{2.4}\cdots\frac{(n+1)^2}{n(n+2)}=\frac{2.(n+1)!^2}{n!.(n+2)!}=\frac{2
(n+1)}{n+2}<2 $