Answer
The Taylor series centered at $c=-1$ is
$T\left( x \right) = - \mathop \sum \limits_{n = 0}^\infty \dfrac{{{3^n}}}{{{5^{n + 1}}}}{\left( {x + 1} \right)^n}$
It is valid for $\left| {x + 1} \right| \lt \dfrac{5}{3}$.
Work Step by Step
We compute the derivatives and list them in the table below:
$\begin{array}{*{20}{c}}
n&{{f^{\left( n \right)}}\left( x \right)}\\
0&{\dfrac{1}{{3x - 2}}}\\
1&{ - \dfrac{3}{{{{\left( {3x - 2} \right)}^2}}}}\\
2&{\dfrac{{18}}{{{{\left( {3x - 2} \right)}^3}}}}\\
3&{ - \dfrac{{162}}{{{{\left( {3x - 2} \right)}^4}}}}\\
4&{\dfrac{{1944}}{{{{\left( {3x - 2} \right)}^5}}}}
\end{array}\begin{array}{*{20}{c}}
{\dfrac{{{f^{\left( n \right)}}\left( x \right)}}{{n!}}}&{\dfrac{{{f^{\left( n \right)}}\left( { - 1} \right)}}{{n!}}}\\
{\dfrac{1}{{3x - 2}}}&{ - \dfrac{1}{5}}\\
{ - \dfrac{3}{{{{\left( {3x - 2} \right)}^2}}}}&{ - \dfrac{3}{{25}}}\\
{\dfrac{9}{{{{\left( {3x - 2} \right)}^3}}}}&{ - \dfrac{9}{{125}}}\\
{ - \dfrac{{27}}{{{{\left( {3x - 2} \right)}^4}}}}&{ - \dfrac{{27}}{{625}}}\\
{\dfrac{{81}}{{{{\left( {3x - 2} \right)}^5}}}}&{ - \dfrac{{81}}{{3125}}}
\end{array}$
By Theorem 1, the Taylor series is given by
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( { - 1} \right)}}{{n!}}{\left( {x + 1} \right)^n}$
Using the table we have obtained above, we get
$T\left( x \right) = - \dfrac{1}{5} - \dfrac{3}{{25}}\left( {x + 1} \right) - \dfrac{9}{{125}}{\left( {x + 1} \right)^2} - \dfrac{{27}}{{625}}{\left( {x + 1} \right)^3} - \dfrac{{81}}{{3125}}{\left( {x + 1} \right)^4} + \cdot\cdot\cdot$
So, $T\left( x \right) = - \mathop \sum \limits_{n = 0}^\infty \dfrac{{{3^n}}}{{{5^{n + 1}}}}{\left( {x + 1} \right)^n}$.
Let ${a_n} = - \dfrac{{{3^n}}}{{{5^{n + 1}}}}{\left( {x + 1} \right)^n}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{3^{n + 1}}}}{{{5^{n + 2}}}}\dfrac{{{{\left( {x + 1} \right)}^{n + 1}}}}{{{{\left( {x + 1} \right)}^n}}}\dfrac{{{5^{n + 1}}}}{{{3^n}}}} \right| = \dfrac{3}{5}\mathop {\lim }\limits_{n - > \infty } \left| {x + 1} \right|$
$\rho = \dfrac{3}{5}\left| {x + 1} \right|$
We find that $\rho \lt 1$ if $\dfrac{3}{5}\left| {x + 1} \right| \lt 1$, or $\left| {x + 1} \right| \lt \dfrac{5}{3}$.
Thus, the Taylor series centered at $c=-1$ is $T\left( x \right) = - \mathop \sum \limits_{n = 0}^\infty \dfrac{{{3^n}}}{{{5^{n + 1}}}}{\left( {x + 1} \right)^n}$. It is valid for $\left| {x + 1} \right| \lt \dfrac{5}{3}$.
