Lời giải:
\(A=\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+...+(\frac{1}{2})^{98}+(\frac{1}{2})^{99}\)
\(\Rightarrow 2A=1+\frac{1}{2}+(\frac{1}{2})^2+...+(\frac{1}{2})^{97}+(\frac{1}{2})^{98}\)
Trừ theo vế:
\(2A-A=1-(\frac{1}{2})^{99}\)
\(A=1-(\frac{1}{2})^{99}< 1\)
Ta có đpcm.