Chứng minh rằng: B = 1/7 + 1/8 + 1/9 + ... + 1/18 + 1/19 < 2

* Hướng dẫn giải

$B=\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\mathrm{...}+\frac{1}{18}+\frac{1}{19}<2$$\iff \left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{7}\right)+\left(\frac{1}{9}-\frac{1}{7}\right)+\mathrm{...}+\left(\frac{1}{18}-\frac{1}{7}\right)+\left(\frac{1}{19}-\frac{1}{7}\right)+\left(\frac{1}{20}-\frac{1}{7}\right)<0$

Vì  $\frac{1}{8}<\frac{1}{7},\frac{1}{9}<\frac{1}{7},\mathrm{...},\frac{1}{20}<\frac{1}{7}$ nên $\frac{1}{8}-\frac{1}{7}<0,\frac{1}{9}-\frac{1}{7}<0,\frac{1}{10}-\frac{1}{7}<0,\mathrm{...},\frac{1}{19}-\frac{1}{7}<0,\frac{1}{20}-\frac{1}{7}<0$

Vậy B < 0