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##