Giải thích các bước giải:
Ta có :
$A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+..+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}$
$\to 3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+..+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}$
$\to 3A+A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+..+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}$
$\to 4A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+..+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}$
$\to -\dfrac 13.4A=-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+..+\dfrac{1}{3^{100}}+\dfrac{100}{3^{101}}$
$\to 4A-(-\dfrac 13.4A)=1-\dfrac{99}{3^{100}}+\dfrac{100}{3^{101}}$
$\to \dfrac{16A}{3}=1-\dfrac{1}{3^{100}}(99-\dfrac{100}{3})<1$
$\to A<\dfrac{3}{16}$
