Question

giúp mình với 

Answer 1

Lời giải:
$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}$

$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}$

$\Rightarrow 2A=3A-A=(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}})-(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}})$

$\Rightarrow 2A=1-\frac{1}{3^{99}}<1$
$\Rightarrow A< \frac{1}{2}$