A power series centered at` c=0` is follows the formula:

`sum_(n=0)^oo a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+...`

The given function `f(x)= 1/(1-3x)` resembles the power series:

`(1+x)^k = sum_(n=0)^oo (k(k-1)(k-2)...(k-n+1))/(n!) x ^n`

or

`(1+x)^k = 1+kx +(k(k-1))/(2!)x^2+(k(k-1)(k-2))/(3!)x^3+(k(k-1)(k-2)(k-3))/(4!)x^4+...`

To evaluate the given function `f(x) =1/(1-3x)` centered at `c=0` , we may apply Law of exponents: `1/x^n = x^(-n)` .

`f(x)= (1-3x) ^(-1)`

Apply the aforementioned formula for power series on `(1-3x) ^(-1) or (1+(-3x))^(-1)` , we may replace "`x` " with "`-3x` " and "k" with "`-1` ". We let:

`(1+(-3x))^(-1) = sum_(n=0)^oo (-1(-1-1)(-1-2)...(-1-n+1))/(n!) (-3x) ^n`

`=sum_(n=0)^oo (-1(-2)(-3)...(-1-n+1))/(n!)(-3)^nx ^n`

`=1+(-1)(-3)^1x +(-1(-2))/(2!)(-3)^2x ^2+(-1(-2)(-3))/(3!)(-3)^3x ^3+(-1(-2)(-3)(-4))/(4!)(-3)^4x ^4+...`

`=1+3x +2/2*9*x ^2+(-6)/6(-27)x ^3+24/24*81*x ^4+...`

`=1+3x +9x ^2+27x ^3+81x ^4+...`

`= sum_(n=0)^oo (3x)^n`

To determine the interval of convergence, we may apply geometric series test wherein the series `sum_(n=0)^oo a*r^n ` is convergent if `|r|lt1 or -1 ltrlt 1` . If `|r|gt=1` then the geometric series diverges.

By comparing `sum_(n=0)^oo (3x)^n` or` sum_(n=0)^oo 1*(3x)^n ` with `sum_(n=0)^oo a*r^n` , we determine: `r =3x` .

Apply the condition for convergence of geometric series: `|r|lt1 ` .

`|3x|lt1`

`-1 lt3xlt1`

Divide each part by 3:

`(-1)/3 lt(3x)/3lt1/3`

`-1/3ltxlt1/3`

Final answer:

The **power series** `sum_(n=0)^oo(3x)^n` has an** interval of convergence**: `-1/3 ltxlt1/3` .