Сравните числа A=199(1+12+⋯+199)A = \frac{1}{{99}}\left( {1 + \frac{1}{2} + \dots + \frac{1}{{99}}} \right)A=991(1+21+⋯+991) и B=1100(1+12+⋯+1100)B = \frac{1}{{100}}\left( {1 + \frac{1}{2} + \dots + \frac{1}{{100}}} \right)B=1001(1+21+⋯+1001)