Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{1}{5} > \dfrac{1}{6} > \dfrac{1}{7} > \dfrac{1}{8}\\
\dfrac{1}{9} > \dfrac{1}{{10}} > \dfrac{1}{{11}} > ..... > \dfrac{1}{{16}}\\
B = \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + .... + \dfrac{1}{{19}}\\
= \dfrac{1}{4} + \left( {\dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8}} \right) + \left( {\dfrac{1}{9} + \dfrac{1}{{10}} + ..... + \dfrac{1}{{16}}} \right) + \left( {\dfrac{1}{{17}} + \dfrac{1}{{18}} + \dfrac{1}{{19}}} \right)\\
> \dfrac{1}{4} + \left( {\dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}} \right) + \left( {\dfrac{1}{{16}} + \dfrac{1}{{16}} + ..... + \dfrac{1}{{16}}} \right) + 0\\
= \dfrac{1}{4} + 4.\dfrac{1}{8} + 8.\dfrac{1}{{16}} = \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{5}{4} > 1\\
\Rightarrow B > 1
\end{array}\)
Vậy \(B > 1\)
