Answer
\[\begin{align}
& \left. a \right)\ln 3+\frac{1}{3}\left( x-3 \right)-\frac{1}{18}{{\left( x-3 \right)}^{2}}+\frac{1}{81}{{\left( x-3 \right)}^{3}} \\
& \left. b \right)\ln 3+\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k+1}}{{\left( x-3 \right)}^{k}}}{k\left( {{3}^{k}} \right)}} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\ln x,\text{ }a=3 \\
& \text{Using the definition of Taylor series for a function}\left( page\,685 \right) \\
& \sum\limits_{k=0}^{\infty }{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}\text{ }\left( \mathbf{1} \right) \\
& =f\left( a \right)+f'\left( a \right)\left( x-a \right)+\frac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\cdots \\
& \text{Calculate some derivatives and their value at }x=3 \\
& f\left( x \right)=\ln x\to f\left( 3 \right)=\ln 3 \\
& f'\left( x \right)=\frac{1}{x}\to f'\left( 3 \right)=\frac{1}{3} \\
& f''\left( x \right)=-\frac{1}{{{x}^{2}}}\to f''\left( 3 \right)=-\frac{1}{9} \\
& f'''\left( x \right)=2{{x}^{-3}}\to {{f}^{\left( 3 \right)}}\left( 3 \right)=\frac{2}{27} \\
& \\
& \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\
& \left( \text{Using the nonzero coefficients} \right) \\
& =\frac{\ln 3}{0!}{{\left( x-3 \right)}^{0}}+\frac{1}{3}{{\left( x-3 \right)}^{1}}+\frac{-1/9}{2!}{{\left( x-3 \right)}^{2}}+\frac{2/27}{3!}{{\left( x-3 \right)}^{3}} \\
& \\
& \text{Simplifying} \\
& =\ln 3+\frac{1}{3}{{\left( x-3 \right)}^{1}}-\frac{1}{18}{{\left( x-3 \right)}^{2}}+\frac{1}{81}{{\left( x-3 \right)}^{3}} \\
& \\
& \left. b \right)\text{ Rewrite the power series} \\
& =\ln 3+\frac{{{\left( -1 \right)}^{1+1}}}{\left( 1 \right){{3}^{1}}}{{\left( x-3 \right)}^{1}}+\frac{{{\left( -1 \right)}^{2+1}}}{\left( 2 \right){{3}^{2}}}{{\left( x-3 \right)}^{2}}+\frac{{{\left( -1 \right)}^{3+1}}}{\left( 3 \right){{3}^{3}}}{{\left( x-3 \right)}^{3}} \\
& \text{Using summation notation}\text{, we obtain} \\
& =\ln 3+\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k+1}}{{\left( x-3 \right)}^{k}}}{k\left( {{3}^{k}} \right)}} \\
\end{align}\]
