chứng tỏ E <3/16
E=1/3-2/3^2+3/3^3-4/3^4+...+2015/3^2015-2016/3^2016
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MV
14 tháng 5 2017
\(E=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 3E=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\\ 3E+E=\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\right)\\ 4E=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 4E< 1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}\left(1\right)\)
Gọi \(D=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{2015}}\)
\(3D=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}\\ 3D+D=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}\right)\\ 4D=3-\dfrac{1}{3^{2015}}< 3\\ \Rightarrow D< \dfrac{3}{4}\left(2\right)\)
Từ (1) và (2) ta có:
\(4E< \dfrac{3}{4}\\ \Rightarrow E< \dfrac{3}{16}\)
27 tháng 7 2017
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
\(E=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 3E=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\\ 3E+E=\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{2015}{3^{2014}}-\dfrac{2016}{3^{2015}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{2015}{3^{2015}}-\dfrac{2016}{3^{2016}}\right)\\ 4E=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 12E=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}-\dfrac{6048}{3^{2016}}\\ 4E+12E=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{2013}}-\dfrac{1}{3^{2014}}-\dfrac{2016}{3^{2015}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}-\dfrac{1}{3^{2015}}-\dfrac{2016}{3^{2016}}\right)\\ 16E=3-\dfrac{2017}{3^{2015}}-\dfrac{2016}{3^{2016}}\\ 16E=3-\left(\dfrac{2017}{3^{2015}}+\dfrac{672}{3^{2015}}\right)\\ 16E=3-\dfrac{2689}{3^{2015}}< 3\\ \Rightarrow E< \dfrac{3}{16}\)