Ta có : \(\dfrac{1}{9}=\dfrac{1}{9}\)
\(\dfrac{1}{10}< \dfrac{1}{9}\)
.....
\(\dfrac{1}{19}< \dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{19}< \dfrac{1}{9}+\dfrac{1}{9}+...+\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{9}+\dfrac{1}{10}+..+\dfrac{1}{19}< \dfrac{11}{9}\)
Hay \(\dfrac{1}{9}+\dfrac{1}{10}+..+\dfrac{1}{19}< \dfrac{9}{9}=1\)
Đặt biểu thức trên là A.
Ta có A có 11 số hạng, chia A thành 2 nhóm, mỗi nhóm có 5 số hạng còn thừa 1 số hạng như sau:
\(A=\dfrac{1}{9}+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}\right)+\left(\dfrac{1}{15}+\dfrac{1}{16}+...+\dfrac{1}{19}\right)\)
Lại có: \(\dfrac{1}{10}=\dfrac{1}{10};\dfrac{1}{11}< \dfrac{1}{10};...;\dfrac{1}{14}< \dfrac{1}{10}\) \(\Rightarrow\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}< \dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}\) (5 số hạng)
\(\Rightarrow\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}< \dfrac{1}{10}.5=\dfrac{1}{2}\) (1)
\(\dfrac{1}{15}=\dfrac{1}{15};\dfrac{1}{16}< \dfrac{1}{15};...;\dfrac{1}{19}< \dfrac{1}{15}\)
\(\Rightarrow\dfrac{1}{15}+\dfrac{1}{16}+...+\dfrac{1}{19}< \dfrac{1}{15}+\dfrac{1}{15}+...+\dfrac{1}{15}\) (5 số hạng)
\(\Rightarrow\dfrac{1}{15}+\dfrac{1}{16}+...+\dfrac{1}{19}< \dfrac{1}{15}.5=\dfrac{1}{3}\)(2)
\(\dfrac{1}{9}=\dfrac{1}{9}\left(3\right)\)
Từ (1) và (2) ta suy ra:
\(\dfrac{1}{9}+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}\right)+\left(\dfrac{1}{15}+\dfrac{1}{16}+...+\dfrac{1}{19}\right)< \dfrac{1}{9}+\dfrac{1}{2}+\dfrac{1}{3}\) \(\Rightarrow A< \dfrac{1}{9}+\dfrac{1}{2}+\dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{2}{18}+\dfrac{9}{18}+\dfrac{6}{18}\)
\(\Rightarrow A< \dfrac{2+9+6}{18}\)
\(\Rightarrow A< \dfrac{17}{18}< \dfrac{18}{18}=1\)
\(\Rightarrow A< 1\left(đpcm\right)\)