Answer
The Taylor series centered at $c=3$ is $T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^{n + 1}}\left( {{2^{n + 1}} - 1} \right)}}{{{2^{2n + 3}}}}{\left( {x - 3} \right)^n}$. It is valid for $\left| {x - 3} \right| \lt 2$.
Work Step by Step
We compute the derivatives and list them in the following table:
$\begin{array}{*{20}{c}}
n&{{f^{\left( n \right)}}\left( x \right)}\\
0&{\dfrac{1}{{1 - {x^2}}}}\\
1&{\dfrac{{2x}}{{{{\left( { - 1 + {x^2}} \right)}^2}}}}\\
2&{ - \dfrac{{2\left( {1 + 3{x^2}} \right)}}{{{{\left( { - 1 + {x^2}} \right)}^3}}}}\\
3&{\dfrac{{24\left( {x + {x^3}} \right)}}{{{{\left( { - 1 + {x^2}} \right)}^4}}}}\\
4&{ - \dfrac{{24\left( {1 + 10{x^2} + 5{x^4}} \right)}}{{{{\left( { - 1 + {x^2}} \right)}^5}}}}
\end{array}\begin{array}{*{20}{c}}
{\dfrac{{{f^{\left( n \right)}}\left( x \right)}}{{n!}}}&{\dfrac{{{f^{\left( n \right)}}\left( 3 \right)}}{{n!}}}\\
{\dfrac{1}{{1 - {x^2}}}}&{ - \dfrac{1}{8}}\\
{\dfrac{{2x}}{{{{\left( { - 1 + {x^2}} \right)}^2}}}}&{\dfrac{3}{{32}}}\\
{\dfrac{{ - 1 - 3{x^2}}}{{{{\left( { - 1 + {x^2}} \right)}^3}}}}&{ - \dfrac{7}{{128}}}\\
{\dfrac{{4\left( {x + {x^3}} \right)}}{{{{\left( { - 1 + {x^2}} \right)}^4}}}}&{\dfrac{{15}}{{512}}}\\
{\dfrac{{ - 1 - 10{x^2} - 5{x^4}}}{{{{\left( { - 1 + {x^2}} \right)}^5}}}}&{ - \dfrac{{31}}{{2048}}}
\end{array}$
By Theorem 1, the Taylor series is given by
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 3 \right)}}{{n!}}{\left( {x - 3} \right)^n}$
Using the table we have obtained, we get
$T\left( x \right) = - \dfrac{1}{8} + \dfrac{3}{{32}}\left( {x - 3} \right) - \dfrac{7}{{128}}{\left( {x - 3} \right)^2} + \dfrac{{15}}{{512}}{\left( {x - 3} \right)^3} - \dfrac{{31}}{{2048}}{\left( {x - 3} \right)^4} + \cdot\cdot\cdot$
So, $T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^{n + 1}}\left( {{2^{n + 1}} - 1} \right)}}{{{2^{2n + 3}}}}{\left( {x - 3} \right)^n}$.
Let ${a_n} = \dfrac{{{{\left( { - 1} \right)}^{n + 1}}\left( {{2^{n + 1}} - 1} \right)}}{{{2^{2n + 3}}}}{\left( {x - 3} \right)^n}$. We compute $\rho $ from the Ratio Test:
$\rho = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\left( {{2^{n + 2}} - 1} \right)}}{{{2^{2n + 5}}}}\dfrac{{{{\left( {x - 3} \right)}^{n + 1}}}}{{{{\left( {x - 3} \right)}^n}}}\dfrac{{{2^{2n + 3}}}}{{{2^{n + 1}} - 1}}} \right|$
$ = \dfrac{1}{4}\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{2^{n + 2}} - 1}}{{{2^{n + 1}} - 1}}\left( {x - 3} \right)} \right| = \dfrac{1}{4}\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{2 - \dfrac{1}{{{2^{n + 1}}}}}}{{1 - \dfrac{1}{{{2^{n + 1}}}}}}\left( {x - 3} \right)} \right|$
$ = \dfrac{1}{2}\left| {x - 3} \right|$
We find that $\rho \lt 1$ if $\dfrac{1}{2}\left| {x - 3} \right| \lt 1$, or $\left| {x - 3} \right| \lt 2$.
Thus, the Taylor series centered at $c=3$ is $T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^{n + 1}}\left( {{2^{n + 1}} - 1} \right)}}{{{2^{2n + 3}}}}{\left( {x - 3} \right)^n}$. It is valid for $\left| {x - 3} \right| \lt 2$.
