a) (1+11.3)(1+12.4)...(1+1n(n+2))<2 Chú ý 1+1n(n+2)=(n+1)2n(n+2). Do đó (1+11.3)(1+12.4)...(1+1n(n+2))=221.3.322.4⋯(n+1)2n(n+2)=2.(n+1)!2n!.(n+2)!=2n+2<2
a)
(1+11.3)(1+12.4)...(1+1n(n+2))<2 Chú ý
1+1n(n+2)=(n+1)2n(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 $