# `sum_(n=0)^oo 3/5^n` Determine the convergence or divergence of the series.

Recall that infinite series converges to single finite value `S`   if the limit if 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 3/5^n` , we may express it in a form:

`sum_(n=0)^oo 3/5^n =sum_(n=0)^oo 3* (1/5^n)`

`=sum_(n=0)^oo 3 *(1/5)^n`

This resembles form of geometric series with an index shift:` sum_(n=0)^oo a*r^n` .

By comparing "`3 *(1/5)^n` " with  "`a*r^n ` ", we determine the corresponding values: `a = 3` and `r =1/5 ` or `0.2` .

The convergence test for the geometric series follows the conditions:

a) If `|r|lt1`  or `-1 ltrlt1 ` then the geometric series converges to `sum_(n=0)^oo a*r^n = a/(1-r)` .

b) If `|r|gt=1` then the geometric series diverges.

The `r=1/5` or `0.2` from the given infinite series falls within the condition `|r|lt1` since `|1/5|lt1` or `|0.2|lt1` . Therefore, we may conclude that `sum_(n=0)^oo 3/5^n` is a convergent series.

By applying the formula: `sum_(n=0)^oo a*r^n= a/(1-r)` , we determine that the given geometric series will converge to a value:

`sum_(n=0)^oo 3/5^n =sum_(n=0)^oo 3 *(1/5)^n`

`= 3/(1-1/5)`

` =3/(5/5-1/5)`

` =3/(4/5)`

` =3*(5/4)`

` = 15/4 or 3.75`

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