nth term is given is 3^-n or 1/3^n. So the first term is 1/3, the second is 1/3^2
and so on.

The series is: 1/3,1/9,1/27... to infinity.

It is an infinite geometric progression with 1st term a as 1/3 and the common 
ratio r as 1/3.

The formula for the sum to infinity of such a progression is given by
S = a / (1 - r) (I'm not including the proof for this here)

So here it is 


The answer is 

:)