Multiply it!

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

$= \frac {1}{3} + \frac {1}{3} ( \frac {1}{3} + \frac {1}{3^{2}} + ... + \frac {1}{3^{7}} )$

$= \frac {1}{3} + \frac {1}{3} ( \frac {1}{3} + \frac {1}{3^{2}} + ... + \frac {1}{3^{7}} + \frac {1}{3^{8}} - \frac {1}{3^{8}} )$

$= \frac {1}{3} + \frac {1}{3} ( A - \frac {1}{3^{8}})$

$\therefore A = \frac {1}{3} + \frac {1}{3} ( A - \frac {1}{3^{8}})$

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

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

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

$\boxed {A = \frac {3280}{6561}}$

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

$\displaystyle \quad = \frac {1}{3} \sum _{n=0}^7 {\left( \frac{1}{3} \right)^n} = \frac {1}{3} \left( \frac{1-\left( \frac{1}{3} \right)^8}{1 - \frac{1}{3}} \right) = \frac {1}{2} \left(\frac {3^8-1} {3^8} \right)$

$\displaystyle \quad = \frac {1}{2} \left(\frac {6561-1} {6561} \right) = \boxed{\dfrac {3280}{6561}}$

A is the sum of the Geometric series with the common ratio (r = 1/3). We can use the below formula to compute the sum

      A = a(1-r^n)/(1-r)

where

a  = first term         =    1/3

r  = common ratio       =    1/3

n  = number of terms    =     8