\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+........+\frac{100}{2^{100}}\)

\(\Rightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+..........+\frac{100}{2^{99}}\)

\(\Rightarrow2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{99}}+\frac{100}{2^{100}}\)

\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{98}}+\frac{100}{2^{99}}\)

\(\Rightarrow2A-A=A=2-\frac{100}{2^{99}}< 2\)

Vậy \(A< 2\)

a)Đặt A= \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{8}\) - \(\frac{1}{16}\) + \(\frac{1}{32}\) - \(\frac{1}{64}\) => A=\(\frac{1}{2^1}\) - \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) - \(\frac{1}{2^4}\) + \(\frac{1}{2^5}\) - \(\frac{1}{2^6}\)

=> 2A= 1-\(\frac{1}{2^1}\) + \(\frac{1}{2^2}\) - \(\frac{1}{2^3}\) + \(\frac{1}{2^4}\) - \(\frac{1}{2^5}\) 

=> 3A= 1- \(\frac{1}{2^6}\) <1 => A<\(\frac{1}{3}\) => đpcm.

b) Đặt B=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\) - \(\frac{4}{3^4}\) +..+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\) 

=> 3B=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\) +...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)

=> 4B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) - \(\frac{100}{3^{99}}\) < 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) (1)

Đặt B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) 

=> 3B= 3-1+\(\frac{1}{3}\) - \(\frac{1}{3^2}\) + \(\frac{1}{3^3}\) - \(\frac{1}{3^4}\) +...+ \(\frac{1}{3^{98}}\)

=> 4B= 3-\(\frac{1}{3^{99}}\) <3 => B<\(\frac{3}{4}\) (2)

=> 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}\)

\(A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)

\(A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)

\(A< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}< \frac{1}{4}+\frac{1}{2}\)

\(A< \frac{1}{4}+\frac{2}{4}=\frac{3}{4}\left(đpcm\right)\)