A geometric series is written as {a,ar,ar^2,ar^3,ar^4,.}, where a is first term (of the series) and r is the ratio of a term to its preceding term.
n^(th) of such a series is ar^(n-1) and sum of the series S_n up to n terms is given by
S_n=a(r^n-1)/(r-1) (used if |r|>1)
or S_n=a(1-r^n)/(1-r) (used if |r|<1)
If |r|<1, sum of infinite series is a/(1-r)
Here a=1/3 and r=(1/9)/(1/3)=1/9xx3/1=1/3
Hence, n^(th) term of the series is 1/3(1/3)^(n-1)=1/3^n and sum of the series S_n is
S_n=1/3((1-(1/3)^n))/((1-1/3))=1/3((1-(1/3^n)))/(2/3)
= 1/3(1-(1/3^n))xx3/2
= 1/2(1-(1/3^n)
And sum of infinite series is 1/2
