# Find a power series representation for the function and determine the interval o

Find a power series representation for the function and determine the interval of convergence.
$$\displaystyle{f{{\left({x}\right)}}}=\frac{{2}}{{3}}-{x}$$

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