Giải:
Ta có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2014^2}>0_{\left(1\right)}.\)(Do S là phân số).
Ta lại có:
\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2014^2}.\)
\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2014.2014}.\)
\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2013.2014}.\)
\(< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}.\)
\(< 1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{2013}-\dfrac{1}{2013}\right)-\dfrac{1}{2014}.\)
\(< 1+0+0+0+...+0-\dfrac{1}{2014}.\)
\(< 1-\dfrac{1}{2014}.\)
\(< \dfrac{2013}{2014}.\)
\(\Rightarrow S< 1_{\left(2\right)}.\) (do \(\dfrac{2013}{2014}< 1\)).
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\) \(\Rightarrow\) \(0< S< 1.\)
\(\Rightarrow S\) không phải là số tự nhiên.
Vậy ta thu được \(đpcm.\)
~ Học tốt!!! ~
