In geometric sequence there is a common ratio, r, between two successive terms.

a2 = a1 *r

a3 = a2 *r = a1 *r^2

....

an = a1 *r^(n-1)

In a geometric series up to infinity, if the absolute value of r is less than 1, then the series converges and its sum is a1/(1-r).

1 +3/5 +9/25 +27/125 +...

r = (3/5)/1 = 3/5

r = (9/25)/(3/5) = (9/25)*(5/3) = 3/5

r = (27/125)/(9/25) = (27/125)*(25/9) = 3/5

Ok, r = 3/5 = 0.6 ----less than 1, so there is a sum.

Sum = a1 / (1-r)

Sum = 1 / (1 -3/5) = 1/(2/5) = 5/2 -------------answer.