Ta có:
\(\dfrac{1}{2}+\dfrac{1}{3}< \dfrac{1}{2}+\dfrac{1}{2}\)
\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}< \dfrac{1}{4}\cdot4\)
\(\dfrac{1}{8}+\dfrac{1}{9}+...+\dfrac{1}{15}< \dfrac{1}{8}\cdot8\)
⇒ \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{15}< \dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{4}\cdot4+\dfrac{1}{8}\cdot8\)
\(\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{31}< \dfrac{1}{16}\cdot16\)
\(\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{63}< \dfrac{1}{32}\cdot32\)
⇒ \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}< 1+\dfrac{1}{2}\cdot2+\dfrac{1}{4}\cdot4+\dfrac{1}{8}\cdot8+\dfrac{1}{16}\cdot16+\dfrac{1}{32}\cdot32\)⇒ \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}< 1+1+1+1+1+1\)
⇒ \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}< 6\)
