Ta có: $A = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \dots + \frac{1}{6561}$

$A = \frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \dots + \frac{1}{3^8}$

$3A = 3 \cdot \left( \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^8} \right)$

$3A = 1 + \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^7}$

$3A - A = 2A = \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^7} \right) - \left( \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^8} \right)$

$2A = 1 - \frac{1}{3^8}$

$2A = 1 - \frac{1}{6561} = \frac{6560}{6561}$

$A = \frac{6560}{6561} : 2 = \frac{3280}{6561}$

Vậy tổng $A = \frac{3280}{6561}$