cho B=[tex]\frac{1}{4}+\frac{1}{5}+\frac{1}{6}... +\frac{1}{19}[/tex] . Hãy chứng minh B>1
ko lấy trên mạng
$ \frac{1}{4} > \frac{1}{8} \\ \frac{1}{5} > \frac{1}{8} \\ ... \\ \frac{1}{7} > \frac{1}{8} \\\Rightarrow \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} > 4 . \frac{1}{8} = \frac{1}{2} \\\frac{1}{9} > \frac{1}{20} \\ \frac{1}{10} > \frac{1}{20} \\ ... \\ \frac{1}{19} > \frac{1}{20} \\\Rightarrow \frac{1}{9} + \frac{1}{10} + ... + \frac{1}{20} > 11 . \frac{1}{20} = \dfrac{11}{20} > \frac{1}{2} \\\Rightarrow B > \dfrac12 + \dfrac12 = 1 $