Trung học cơ sở

Bài toán: So sánh B với 1 biết: $ \displaystyle B=\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\ldots +\dfrac{1}{{17}}$ Giải: Ta có: $ \displaystyle \dfrac{1}{5}>\dfrac{1}{{10}};\,\,\dfrac{1}{6}>\dfrac{1}{{10}};\,\,\dfrac{1}{7}>\dfrac{1}{{10}};\,\,\dfrac{1}{8}>\dfrac{1}{{10}};\,\,\dfrac{1}{9}>\dfrac{1}{{10}}$ ⇒ $ \displaystyle \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\,\dfrac{1}{8}+\dfrac{1}{9}>\dfrac{5}{{10}}=\dfrac{1}{2}$  (1) $…