Đặt A= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
=\(\frac{1}{2^1}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
=> 2A= \(1-\frac{1}{2^1}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
Ta có:
2A+A=\(\left(1-\frac{1}{2^1}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2^1}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
=> 3A=\(1-\frac{1}{2^1}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}+\frac{1}{2^1}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
=\(1-\left(\frac{1}{2^1}-\frac{1}{2^1}\right)+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)-\left(\frac{1}{2^3}-\frac{1}{2^3}\right)+\left(\frac{1}{2^4}-\frac{1}{2^4}\right)-\left(\frac{1}{2^5}-\frac{1}{2^5}\right)-\frac{1}{2^6}\)
= \(1-\frac{1}{2^6}\)
=> A= 3A:3= \(\left(1-\frac{1}{2^6}\right):3\)=\(\frac{1}{3}-\frac{1}{2^6}:3\)<\(\frac{1}{3}\)
