sum_(n=0)^oo 3^n/1000 Determine the convergence or divergence of the series.

Recall that infinite series converge to a single finite value S  if the limit of the partial sum S_n as n approaches oo converges to S . We follow it in a formula:
lim_(n-gtoo) S_n=sum_(n=1)^oo a_n = S .
To evaluate the  sum_(n=0)^oo 3^n/1000 , we may express it in a form:
sum_(n=0)^oo 1/1000 * 3^n .
 This resembles form of geometric series with an index shift:  sum_(n=0)^oo a*r^n .
By comparing "1/1000 * 3^n " with  "a*r^n ", we determine the corresponding values: a = 1/1000 and r = 3 .
 The convergence test for the geometric series follows the conditions:
 a) If |r|lt1  or -1 ltrlt 1 then the geometric series converges to sum_(n=0)^oo a*r^n =sum_(n=1)^oo a*r^(n-1)= a/(1-r) .
 b) If |r|gt=1 then the geometric series diverges.
The r=3 from the given infinite series falls within the condition |r|gt=1 since |3|gt=1 . Therefore, we may conclude that sum_(n=0)^oo 3^n/1000  is a divergent series.