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Chứng minh rằng :
\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{2014}}\) \(< \frac{1}{2}\)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\)=>\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2013}}\)
=>\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2013}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}\right)\)
=>\(2A=1-\frac{1}{2^{2014}}< 1\Rightarrow A< \frac{1}{2}\)(đpcm)
\(M=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{10}{3^{11}}\)
\(\Rightarrow3M=\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\)
\(\Rightarrow3M-M=\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+...+\frac{10}{3^{11}}\right)\)
\(\Rightarrow2M=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}-\frac{10}{3^{11}}\)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\)
\(\Rightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^9}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^9}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\right)\)
\(\Rightarrow2A=1-\frac{1}{3^{10}}< 1\)
\(\Rightarrow2A< 1\)
\(\Rightarrow A< \frac{1}{2}\)
\(\Rightarrow2M< \frac{1}{2}-\frac{10}{3^{11}}\)
\(\Rightarrow M< \frac{\frac{1}{2}-\frac{10}{3^{11}}}{2}\)
\(\Rightarrow M< \frac{1}{4}-\frac{1}{2.3^{11}}< \frac{1}{4}\)
\(\Rightarrow M< \frac{1}{4}\left(đpcm\right)\)
Ta có: \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
Tương tự : \(\frac{1}{3^2}< \frac{1}{2.3}\); \(\frac{1}{4^2}< \frac{1}{3.4}\); ......... ; \(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{2013.2014}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2013}-\frac{1}{2014}\)
\(=1-\frac{1}{2014}=\frac{2013}{2014}\)
\(\Rightarrow S< \frac{2013}{2014}\left(đpcm\right)\)
\(\frac{9}{10!}+\frac{10}{11!}+...+\frac{999}{1000!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+...+\frac{1000-1}{1000!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+...+\frac{1}{999!}-\frac{1}{1000!}\)
\(=\frac{1}{9!}-\frac{1}{1000!}< \frac{1}{9!}\)
đpcm
Tham khảo nhé~
Đặt \(S=\frac{1}{10^2}+\frac{1}{11^2}+\frac{1}{12^2}+.....+\frac{1}{2014^2}\)
Ta có : \(S< \frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+.....+\frac{1}{2013.2014}\\\)
Đặt \(A=\frac{1}{9.10}+\frac{1}{10.11}+....+\frac{1}{2013.2014}\\ =>A=\left(\frac{1}{9}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{11}\right)+......+\left(\frac{1}{2013}-\frac{1}{2014}\right)\\ =>A=\frac{1}{9}-\frac{1}{2014}\\ \)
Vậy A<\(\frac{1}{9}\)
Mà A>S =>S<\(\frac{1}{9}\)