Explicație pas cu pas:
a)
[tex]\frac{1}{8} < \frac{1}{3} \leqslant \frac{1}{3} \\ \frac{1}{8} < \frac{1}{4} < \frac{1}{3} \\\frac{1}{8} < \frac{1}{5} < \frac{1}{3} \\\frac{1}{8} < \frac{1}{6} < \frac{1}{3} \\\frac{1}{8} < \frac{1}{7} < \frac{1}{3} \\\frac{1}{8} \leqslant \frac{1}{8} < \frac{1}{3} \\[/tex]
adunăm, pe verticală:
[tex]\frac{6}{8} < \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} < \frac{6}{3} \\ [/tex]
[tex]\frac{3}{4} < \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} < 2 \\ [/tex]
b)
[tex]\frac{1}{27} < \frac{1}{10} \leqslant \frac{1}{10} \\ \frac{1}{27} < \frac{1}{11} < \frac{1}{10} \\\frac{1}{27} < \frac{1}{12} < \frac{1}{10} \\ ... \\\frac{1}{27} < \frac{1}{26} < \frac{1}{10} \\\frac{1}{27} \leqslant \frac{1}{27} < \frac{1}{10} \\[/tex]
adunăm, pe verticală:
[tex]\frac{18}{27} < \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + ... + \frac{1}{27} < \frac{18}{10} \\ [/tex]
[tex]\frac{2}{3} < \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + ... + \frac{1}{27} < \frac{9}{5} \\ [/tex]
