\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow3A+A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow4A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow4A+12A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}< 3\)
\(\Rightarrow16A< 3\)
\(\Rightarrow A< \dfrac{3}{16}< \dfrac{3}{6}\)
