Giải thích các bước giải:
$B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}\\
\Rightarrow 2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}\\
\Rightarrow B=2B-B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}-\left ( \dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}} \right )\\
=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}- \dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-\dfrac{1}{2^4}-...-\dfrac{1}{2^{99}} \\
=1-\dfrac{1}{2^{99}}<1\\
b)
C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\\
\Rightarrow 3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\\
\Rightarrow 2C=3C-C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\left ( \dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}} \right )\\
=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{99}}\\
=1-\dfrac{1}{3^{99}}\\
\Rightarrow C=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}<\dfrac{1}{2}$
