Answer
The function: $\dfrac{1}{x}$
The expansion is valid for $\left| {x - 3} \right| \lt 3$.
Work Step by Step
Write
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{3^{k + 1}}}}{\left( {x - 3} \right)^k} = \dfrac{1}{3}\mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( {\dfrac{{x - 3}}{3}} \right)^k}$
$ = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}\dfrac{1}{3}{\left( {\dfrac{x}{3} - 1} \right)^k} = \mathop \sum \limits_{k = 0}^\infty {\left( { - 1} \right)^k}{\left( { - 1} \right)^k}\dfrac{1}{3}{\left( {1 - \dfrac{x}{3}} \right)^k}$
Since ${\left( { - 1} \right)^k}{\left( { - 1} \right)^k} = {\left( { - 1} \right)^{2k}} = 1$, we have a simplified series
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{3^{k + 1}}}}{\left( {x - 3} \right)^k} = \mathop \sum \limits_{k = 0}^\infty \dfrac{1}{3}{\left( {1 - \dfrac{x}{3}} \right)^k}$
Let $t = 1 - \dfrac{x}{3}$, so
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{3^{k + 1}}}}{\left( {x - 3} \right)^k} = \dfrac{1}{3}\mathop \sum \limits_{k = 0}^\infty {t^k}$
From Table 2, we know that $\mathop \sum \limits_{k = 0}^\infty {t^k} = \dfrac{1}{{1 - t}}$, valid for $\left| t \right| \lt 1$.
Therefore,
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{3^{k + 1}}}}{\left( {x - 3} \right)^k} = \dfrac{1}{3}\dfrac{1}{{1 - t}}$
Substituting $t = 1 - \dfrac{x}{3}$ into the right-hand side gives
$\mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{3^{k + 1}}}}{\left( {x - 3} \right)^k} = \dfrac{1}{3}\dfrac{1}{{1 - \left( {1 - \dfrac{x}{3}} \right)}} = \dfrac{1}{x}$
Therefore, the function $\dfrac{1}{x}$ has Maclaurin series $\mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{3^{k + 1}}}}{\left( {x - 3} \right)^k}$.
Since $\left| t \right| \lt 1$, so $\left| {1 - \dfrac{x}{3}} \right| \lt 1$.
Thus,
$\left| {3 - x} \right| \lt 3$ ${\ \ \ }$ or ${\ \ \ }$ $\left| {x - 3} \right| \lt 3$
Thus, the expansion is valid for $\left| {x - 3} \right| \lt 3$.
