Giải thích các bước giải:
a) $S= \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}$
Có: $\frac{1}{3}+\frac{1}{4}> \frac{1}{4}+\frac{1}{4}= \frac{1}{2}$
$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}> \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}= \frac{1}{2}$
$\frac{1}{9}+...+\frac{1}{16} > 8\cdot \frac{1}{16}= \frac{1}{2} $
$\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}> 16\cdot \frac{1}{32}= \frac{1}{2}$
$\Rightarrow S> \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}= \frac{5}{2}$ (đpcm)
b) $S= \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}$
$\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3\cdot \frac{1}{3}$
$\frac{1}{6}+...+\frac{1}{11}< 6\cdot \frac{1}{6}$
$\frac{1}{12}+...+\frac{1}{23}< 12\cdot \frac{1}{12}$
$\frac{1}{24}+...+\frac{1}{32}< 9\cdot \frac{1}{24}$
$\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}= \frac{31}{8}< \frac{9}{2}$(đpcm)
