Ta có:
A = $\frac{1}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$ + ... + $\frac{1}{17}$
= ($\frac{1}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$) + ($\frac{1}{6}$) + $\frac{1}{7}$ + $\frac{1}{8}$) + ... + ($\frac{1}{15}$ + $\frac{1}{16}$ + $\frac{1}{17}$)
=> A < ($\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ ) + ($\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$) + ... + ($\frac{1}{15}$ + $\frac{1}{15}$ + $\frac{1}{15}$)
=> A < 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$
=> A < $\frac{137}{60}$ < 3
Vậy A<3.
