Answer
\[\begin{align}
& \left. a \right)\frac{1}{4}-\frac{1}{16}{{x}^{2}}+\frac{1}{64}{{x}^{4}} \\
& \left. b \right)\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{{{\left( 4 \right)}^{k+1}}}{{x}^{k}}} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\frac{1}{4+{{x}^{2}}},\text{ }a=0 \\
& \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{ } \\
& =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{ }\left( \mathbf{1} \right) \\
& \text{First}\text{, we calculate some derivatives and their value at }x=0 \\
& f\left( x \right)=\frac{1}{4+{{x}^{2}}}\to f\left( 0 \right)=\frac{1}{4+{{\left( 0 \right)}^{2}}}=\frac{1}{4} \\
& f'\left( x \right)=-{{\left( 4+{{x}^{2}} \right)}^{-2}}\left( 2x \right)=-\frac{2x}{{{\left( 4+{{x}^{2}} \right)}^{2}}}\to f'\left( 0 \right)=0 \\
& \text{Differentiating by hand }f''\left( x \right),f'''\left( x \right)\text{, }{{f}^{\left( 4 \right)}}\left( x \right)\text{and replacing} \\
& f''\left( x \right)=\frac{6{{x}^{2}}-8}{{{\left( 4+{{x}^{2}} \right)}^{3}}}\to f'''\left( 0 \right)=\frac{-8}{{{\left( 4 \right)}^{3}}}=-\frac{1}{8} \\
& f'''\left( x \right)=-\frac{24x\left( {{x}^{2}}-4 \right)}{{{\left( 4+{{x}^{2}} \right)}^{4}}}\to {{f}^{\left( 4 \right)}}\left( 0 \right)=0 \\
& {{f}^{\left( 4 \right)}}\left( x \right)=\frac{24\left( 5{{x}^{4}}-40{{x}^{2}}+16 \right)}{{{\left( 4+{{x}^{2}} \right)}^{5}}}\to {{f}^{\left( 4 \right)}}\left( 0 \right)=\frac{24\left( 16 \right)}{{{\left( 4 \right)}^{5}}}=\frac{3}{8} \\
& \text{Then}\text{,} \\
& f\left( 0 \right)=\frac{1}{4},\text{ }f'\left( 0 \right)=0,\text{ }f''\left( 0 \right)=-\frac{1}{8},\text{ }f'''\left( 0 \right)=0 \\
& {{f}^{\left( 4 \right)}}\left( 0 \right)=\frac{3}{8} \\
& \\
& \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\
& \left( \text{Using the nonzero coefficients} \right) \\
& =\frac{1/4}{0!}{{x}^{0}}-\frac{1/8}{2!}{{x}^{2}}+\frac{3/8}{4!}{{x}^{4}} \\
& \text{Simplifying} \\
& =\frac{1}{4}-\frac{1}{16}{{x}^{2}}+\frac{1}{64}{{x}^{4}} \\
& \\
& \left. b \right)\text{ Rewrite the power series} \\
& ={{\left( -1 \right)}^{0}}\frac{1}{{{4}^{0+1}}}{{x}^{0}}+{{\left( -1 \right)}^{1}}\frac{1}{{{\left( 4 \right)}^{1+1}}}{{x}^{2}}+{{\left( -1 \right)}^{2}}\frac{1}{{{\left( 4 \right)}^{2+1}}}{{x}^{4}} \\
& \text{Using summation notation}\text{, we obtain} \\
& =\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{{{\left( 4 \right)}^{k+1}}}{{x}^{k}}} \\
\end{align}\]
