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**.

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