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Đặt: \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.....\frac{2013}{2014}\) (1)
Ta thấy \(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}.....\frac{2014}{2015}\)
Do đó nhân vế với vế, ta được:
\(A^2< \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2013}{2014}.\frac{2014}{2015}\)
\(\Rightarrow A^2< \frac{1}{2015}\)
Mặt khác, \(A>\frac{1}{2}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}.....\frac{2014}{2015}\) (2)
Từ (1) và (2), ta được:
\(A^2>\frac{1}{4}.\left(\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2013}{2014}.\frac{2014}{2015}\right)\)
\(\Rightarrow A^2>\frac{1}{4}.\frac{3}{2015}\Rightarrow A^2>\frac{3}{8060}>\frac{1}{4028}\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(S=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Rightarrow\left(S-P\right)^{2016}=\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}-\frac{1}{1008}-\frac{1}{1009}-...-\frac{1}{2015}\right)^{2016}=0^{2016}=0\)
Ta thấy:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)+\frac{1}{2015}\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)+\frac{1}{2015}\)
\(S=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\)
Mà \(P=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}+\frac{1}{2015}\) nên:
\(S=P\)\(\Rightarrow S-P=0\)\(\Rightarrow\left(S-P\right)^{2016}=0\)
\(\frac{1}{4028}< \frac{1}{2}.....\frac{2013}{2014}< \frac{1}{2015}\)
Xét tích: \(\frac{1}{2}.....\frac{2013}{2014}\) \(\Rightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2013}{2014}\)\(=\frac{1.2.3...2013}{2.3.4...2014}\)\(=\frac{1}{2014}\)
\(\Rightarrow\frac{1}{4028}< \frac{1}{2014}< \frac{1}{2015}\)( Vô lí )