Răspuns:
Folosim [tex]\frac{1}{k\cdot (k+1)}=\frac{k+1-k}{k\cdot (k+1)}=\frac{1}{k}+\frac{1}{k+1}[/tex]
Atunci a = [tex] \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \cdots +
\frac{1}{17}-\frac{1}{18} = \frac{1}{3} - \frac{1}{18} = \frac{5}{18} = 0,2(7) [/tex]
Rezulta ca $a\in [0,(2) ; 0,(3)]$.
