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...
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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