Giải thích các bước giải:
Ta có :
$A=-\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+..+\dfrac{1}{3^{100}}$
$\to 3A=-1+\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^{99}}$
$\to 3A-A=-1+\dfrac{2}{3}-\dfrac{1}{3^{100}}$
$\to 2A=-\dfrac{1}{3}-\dfrac{1}{3^{100}}$
$\to |2A|=\dfrac{1}{3}+\dfrac{1}{3^{100}}<\dfrac 13+\dfrac 16=\dfrac 12\to |A|<\dfrac 14$