chứng minh rằng: [tex]\frac{1}{2}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< 2[/tex]
[tex]\frac{1}{2}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\\\\ =\frac{1}{2}+(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{11})+(\frac{1}{12}+...+\frac{1}{17})\\\\ <\frac{1}{2}+(\frac{1}{6}+\frac{1}{6}+...+\frac{1}{6})+(\frac{1}{12}+\frac{1}{12}+...+\frac{1}{12})\\\\ =\frac{1}{2}+1+\frac{1}{2}=2[/tex]