\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{2015}}-\dfrac{1}{2^{2016}}\)
`=>` \(2A=1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{1}{2^{2017}}\)
`=>` \(2A+A=\left(1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{1}{2^{2017}}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{2015}}-\dfrac{1}{2^{2016}}\right)\)
`=>` \(3A=1-\dfrac{1}{2^{2017}}=\dfrac{2^{2017}-1}{2^{2017}}< 1\)
`=>` \(A< \dfrac{1}{3}\)
Vậy \(A< \dfrac{1}{3}\)
\(A=\dfrac{1}{2^1}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{2015}}-\dfrac{1}{2^{2016}}\)
\(2A=1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}-\dfrac{1}{2^{2015}}\)
`=>3A=A+2A`
`=1/2-1/2^2+1/2^3-1/2^4+..+1/2^2015-1/2^2016+1-1/2+1/2^2-..-1/2^2015`
`=1-1/2^2016 = (2^2016-1)/(2^2016)<1`
`=> 3A <1`
`=> A<1/3`
