\(S=\dfrac{1}{2018}\left(1+\dfrac{1}{1}+1+\dfrac{1}{2}+1+\dfrac{1}{3}+...+1+\dfrac{1}{2018}\right)\)
\(S=\dfrac{1}{2018}\left(2018+\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)
\(S=1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)
Do \(\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2018}\right)>0\Rightarrow S>1\) (1)
Lại có:
\(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}< \dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+...+\dfrac{1}{1}=2018\)
\(\Rightarrow1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)< 1+\dfrac{1}{2018}.2018=2\)
\(\Rightarrow S< 2\) (2)
Từ (1), (2) \(\Rightarrow1< S< 2\)
\(\Rightarrow S\) nằm giữa 2 số tự nhiên liên tiếp nên S không phải là số tự nhiên
