Đặt $A'=\dfrac{1}{3}-\dfrac{2}{3^2}-\dfrac{3}{3^3}-\dfrac{4}{3^4}-...-\dfrac{99}{3^99}-\dfrac{100}{3^{100}}$
$⇒3A=\dfrac{3.1}{3}+\dfrac{2.3}{3^2}+\dfrac{3.3}{3^3}+\dfrac{3.4}{3^4}+...+\dfrac{3.99}{3^{99}}+\dfrac{100.3}{3^{100}}$
$=\dfrac{1}{1}+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}+\dfrac{100}{3^{99}}$
⇒$3A-A=\dfrac{1}{1}+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}+\dfrac{100}{3^{99}}-\dfrac{1}{3}-\dfrac{2}{3^2}-\dfrac{3}{3^3}-\dfrac{4}{3^4}-...-\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}$
⇒$2A=1+(\dfrac{2}{3}+\dfrac{1}{3})+(\dfrac{3}{3^2}+\dfrac{2}{3^2})+(\dfrac{4}{3^3}-\dfrac{3}{3^3})+(\dfrac{5}{3^4}-\dfrac{4}{3^4})+...+(\dfrac{99}{3^{98}}-\dfrac{98}{3^{98}}+(\dfrac{100}{3^{99}}-\dfrac{99}{3^{99}})-\dfrac{100}{3^{100}}$
$=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}$
Đặt $B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}$
⇒$3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}$
⇒$2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-1-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-\dfrac{1}{3^4}-...-\dfrac{1}{3^{99}}$
⇒$2B=1-\dfrac{1}{3^{99}}$
⇒$B=\dfrac{1-\dfrac{1}{3^{99}}}{2}=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}$
Thay vào 2A ta được
$2A=1+\dfrac{1}{2}-\dfrac{1}{2.3^{99}}-\dfrac{100}{3^{100}}$
⇒$A=\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^2.3^{99}}-\dfrac{100}{2^2.3^{100}}<\dfrac{3}{4}$
$A<\dfrac{3}{4}$
