How do you find the sum of the infinite geometric series 1/3+1/9+1/27+1/81+...?

##Soo= 1/2##

Formula for sum of infinite geometric series is ##S_oo=a_1/(1-r)## ; ##" " " " " -1 < r < 1##

We have a geometric series :##1/3 + 1/9 + 1/81+.........##

First we know ##a_1= 1/3## (the first term)

Second: Identify ##r## , we know ##r= a_2/a_1## or ##r= a_n/a_(n-1##

##r= (1/(9))/(1/3)## ##hArr 1/9 *3/1 = 1/3 ##

##r= 1/3##

Substitute into the formula ##Soo= (1/3)/(1-1/3) ##

##= (1/3) /(2/3) ##

##=(1/3)*(3/2)##

##Soo= 1/2##