Xét số hạng tổng quát:
[math]\frac{1}{1+3+5+7+...+(2n-1)}=\frac{1}{\frac{(2n-1+1)n}{2}}=\frac{2}{2n.n}=\frac{1}{n^2}<\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2}(\frac{1}{n-1}-\frac{1}{n+1})[/math]Do đó:
[math]A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+...+2017}[/math][math]\implies A<\frac{1}{1+3}+\frac{1}{2}(\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{1008}-\frac{1}{1010})[/math][math]\implies A<\frac{1}{4}+\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{1009}-\frac{1}{1010})[/math][math]\implies A<\frac{1}{4}+\frac{1}{2}(\frac{1}{2}+\frac{1}{3})=\frac{1}{4}+\frac{1}{4}+\frac{1}{6}=\frac{3+3+2}{12}=\frac{8}{12}<\frac{9}{12}=\frac{3}{4}[/math]Suy ra đpcm.
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
