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Đặt \(S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
Biến đổi mẫu
\(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)
\(=\left(2017+1\right)+\left(\frac{2016}{2}+1\right)+...+\left(\frac{1}{2017}+1\right)-2017\)
\(=2018+\frac{2018}{2}+...+\frac{2018}{2017}+\frac{2018}{2018}-2018\)
\(=2018.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)
\(\Rightarrow S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}{2018.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}=\frac{1}{2018}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
Ta thấy \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}=1+A-\frac{1}{2^{2016}}\)
\(\Rightarrow A=1-\frac{1}{2^{2016}}< 1\)
Vậy A < 1.
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
Ta có :
\(S=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2-1}\right)...\left(\dfrac{1}{2016^2-1}\right)\)
\(S=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{2.4}{3^2}\right)...\left(\dfrac{2015.2017}{2016^2}\right)\)
\(S=\dfrac{1.3.2.4....2015.2017}{2^2.3^2....2016^2}\)
\(S=\dfrac{1.2017}{2.2016}=\dfrac{2017}{4032}\)
⇒ S > -1/2
cảm ơn bạn nha