Question

B = \(\dfrac{1}{2}+\dfrac{1}{^{ }2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^{2016}}\)

CMR B < 1

Answer 1

\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\)

\(\Rightarrow\dfrac{B}{2}=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\)

\(\Rightarrow\dfrac{B}{2}-B=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2017}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\right)\)

\(\Leftrightarrow-\dfrac{B}{2}=\dfrac{1}{2^{2017}}-\dfrac{1}{2}\)

\(\Rightarrow B=1-\dfrac{1}{2^{2016}}< 1\)