A=\(\frac{1}{2}.\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+.......+\frac{1}{9240}\right)>\frac{57}{462}\)ta nhóm 6 là ngoài
A=\(\frac{1}{2}.6.\left(1+\frac{1}{4}+\frac{1}{10}+.........+\frac{1}{9240}\right)>\frac{57}{462}\)
A=3.(1+\(\frac{1}{1540}\))
A=3.\(\frac{1541}{1540}\)
A=3
\(\Rightarrow\)3>\(\frac{1541}{1540}\)