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Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)
Ta có:\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)\(>B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}\)
Mà \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}< 1\)
=>Đpcm
Cách 2:
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + …. + 11x12x(13-10) + 12x13x(14-11)
S x 3 = 1x2x3 + 2x3x4 – 2x3x1 + 3x4x5 – 3x4x2 + …..+ 11x12x13 – 11x12x10 +12x13x14 – 12x13x11
S x 3 = 12 x 13 x14
S = 4 x 13 x 14
S = 728
\(\frac{1}{2^2}+\)\(\frac{1}{3^2}+\)\(\frac{1}{4^2}+\)...+\(\frac{1}{2015^2}+\)\(\frac{1}{2015}\)
<\(\frac{1}{1.2}+\)\(\frac{1}{3.4}+\)\(\frac{1}{4.5}+\)...+\(\frac{1}{2014.2015}\)+\(\frac{1}{2015}\)
Ta có:\(\frac{1}{1.2}+\)\(\frac{1}{3.4}+\)\(\frac{1}{4.5}+\)...+\(\frac{1}{2014.2015}\)+\(\frac{1}{2015}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}\)
=1
=>\(\frac{1}{2^2}+\)\(\frac{1}{3^2}+\)\(\frac{1}{4^2}+\)...+\(\frac{1}{2015^2}+\)\(\frac{1}{2015}\) \(
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2016^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(A< 1-\frac{1}{2016}\)
\(A< \frac{2015}{2016}\left(đpcm\right)\)
\(A=\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{2016.2016}< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
\(\Rightarrow A< \frac{2015}{2016}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)
\(< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(< 1-\frac{1}{2016}< 1\left(đpcm\right)\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\left(\dfrac{2015}{2}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}=\dfrac{1}{2017}\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..........+\frac{1}{2015^2}\)
\(\Leftrightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2014.2025}\)
\(\Leftrightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2014.2015}\)
\(\Leftrightarrow B< 1-\frac{1}{2015}< 1\)
\(\Leftrightarrow B< 1\rightarrowđpcm\)
Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\)
+ Xét : \(\frac{1}{1\cdot2}>\frac{1}{2^2}\)
\(\frac{1}{2\cdot3}>\frac{1}{3^2}\)
\(\frac{1}{3\cdot4}>\frac{1}{4^2}\)
...
\(\frac{1}{2015^2}< \frac{1}{2014\cdot2015}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(A=1-\frac{1}{2015}< 1\)
\(\Rightarrow B< A< 1\left(đpcm\right)\)
Vì \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};...;\frac{1}{2015^2}< \frac{1}{2014\cdot2015}\)
\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2014\cdot2015}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}< 1\)
Vậy \(A< 1\left(đpcm\right)\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2015^2}< \frac{1}{2014.2015}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}< 1^{\left(đpcm\right)}\)
Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\)(Tự chứng minh)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2015}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015}\)
Gọi \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015}\)
Ta có: \(A=\frac{1}{1}-\frac{1}{2015}+\frac{1}{2015}=1\)
Do đó \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2015}<1\)