Answer
$f\left( x \right) = \ln \dfrac{x}{2} = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{{\left( {x - 2} \right)}^n}}}{{{2^n}n}}$
Work Step by Step
Write $\ln \dfrac{x}{2} = \ln \left( {1 + \dfrac{1}{2}\left( {x - 2} \right)} \right)$.
From Table 2 of Section 11.7, we know that
$\ln \left( {1 + x} \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{x^n}}}{n} = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \lt 1$ and $x=1$.
Substituting $\dfrac{1}{2}\left( {x - 2} \right)$ for $x$ in the series above gives
$\ln \left( {1 + \dfrac{1}{2}\left( {x - 2} \right)} \right) = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{{\left( {\dfrac{1}{2}\left( {x - 2} \right)} \right)}^n}}}{n} = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{{\left( {x - 2} \right)}^n}}}{{{2^n}n}}$
Thus, the Taylor series centered at $c=2$ is
$f\left( x \right) = \ln \dfrac{x}{2} = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^{n - 1}}{{\left( {x - 2} \right)}^n}}}{{{2^n}n}}$
