Ta có $\frac1{17} < \frac1{16},\frac1{15}< \frac 1{14},...$$\Rightarrow \frac 19+\frac 1{11}+\frac 1{13}+\frac{1}{15}+ \frac{1}{17} < \frac 1{8} +\frac 1{10} + \frac 1{12} + \frac 1{14}+ \frac 1{16}$
$\Rightarrow \frac 18+\frac 19+\frac 1{10}+...+\frac 1{17}<2 (\frac 1{8} +\frac 1{10} + \frac 1{12} + \frac 1{14}+ \frac 1{16})$
$\Rightarrow \frac 18+\frac 19+\frac 1{10}+...+\frac 1{17} < \frac 14+\frac 15 + \frac 16 + \frac 17 +\frac 18$
$\Rightarrow \frac 14+ \frac 15 +\frac 16+...+ \frac 1{17}<2(\frac 14+\frac 15+\frac 16+ \frac 17)+\frac 18<2.1 +\frac 18$
$\Rightarrow A< \frac 12+\frac 13+2+ \frac 18<3$
Lại có
$A> \frac 12+\frac 13+\frac 14+\frac 15+\frac 16 + \frac 17 + \frac 18+\frac 1{10}+ \frac 1{12}+\frac 1{14} +\frac 1{15} + \frac 1{16}$
$=(\frac 12+ \frac 13+ \frac 1{10} + \frac 1{15})+(\frac 15 +\frac 17)+ (\frac 1{14}+ \frac{1}{16})+( \frac 14+\frac 16 + \frac 18+ \frac 1{12})$
$>1+(\frac 18 + \frac 18)+(\frac 1{16} + \frac 1{16})+ \frac 58 =2$
Vậy ta có $2<A<3$ nên $A$ ko là số tự nhiên