\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{2}}}{{{3}-{x}}}}\)

Divide Numerator and denominator by 3. To get

\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{\frac{{{2}}}{{{3}}}}}}{{{1}-{\frac{{{x}}}{{{3}}}}}}}\)

Note that the sum of the geometric series with Initial term a and common ratio r is

\(\displaystyle{S}={\sum_{{{n}={0}}}^{\infty}}{a}{r}^{{n}}={\frac{{{a}}}{{{1}-{r}}}}\)

The given function can be interpred as

\(\displaystyle{\frac{{{\frac{{{2}}}{{{3}}}}}}{{{1}-{\frac{{{x}}}{{{3}}}}}}}={\frac{{{a}}}{{{1}-{r}}}}\)

Therefore, we can say that f(x) is a sum of a geometric series with initial term \(\displaystyle{a}={\frac{{{2}}}{{{3}}}}\) and common ratio \(\displaystyle{r}={\frac{{{x}}}{{{3}}}}\)

Therefore,

\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{a}{r}^{{n}}={\sum_{{{n}={0}}}^{\infty}}{\left({\frac{{{2}}}{{{3}}}}\right)}{\left({\frac{{{x}}}{{{3}}}}\right)}^{{n}}={\sum_{{{n}={0}}}^{\infty}}{\left({\frac{{{2}}}{{{3}^{{{n}+{1}}}}}}\right)}\cdot{x}^{{n}}\)

This is the power series representation of f(x)

We know that the geometric series converges when \(\displaystyle{\left|{r}\right|}={\left|{\frac{{{x}}}{{{3}}}}\right|}{<}{1}\)

\(\displaystyle{\left|{x}\right|}{<}{3}\)

Intterval of converges is \(\displaystyle{\left(-{3},{3}\right)}\)

Radius of converges is 3