詳解：\(f(x)\) 在 \(x = 0\) 之馬克勞林級數展開式

\(f(x) \buildrel\textstyle.\over= f(0) + \frac{{f'(0)}}{{1!}}x + \frac{{f''(0)}}{{2!}}{x^2} + \frac{{f'''(0)}}{{3!}}{x^3} + \frac{{{f^{(4)}}(0)}}{{4!}}{x^4} +  \cdots \)

\(f(x) = \ln (x + 1)\) \(  \Rightarrow f(0) = \ln (1) = 0\)

\(f'(x) = \frac{1}{{x + 1}}\) \(  \Rightarrow f'(0) = \frac{1}{1} = 1\)

\(f''(x) =  - \frac{1}{{{{\left( {x + 1} \right)}^2}}}\) \(  \Rightarrow f''(0) =  - \frac{1}{1} =  - 1\)

\(f'''(x) = \frac{2}{{{{\left( {x + 1} \right)}^3}}}\) \(  \Rightarrow f'''(0) = \frac{2}{1} = 2\)

\({f^{(4)}}(x) = \frac{{ - 3!}}{{{{\left( {x + 1} \right)}^4}}}\) \(  \Rightarrow {f^{(4)}}(0) = \frac{{ - 3!}}{1} =  - 3!\)

\(\ln \left( {x + 1} \right) \buildrel\textstyle.\over= \frac{1}{{1!}}x + \frac{{ - 1}}{{2!}}{x^2} + \frac{2}{{3!}}{x^3} + \frac{{ - 3!}}{{4!}}{x^4} +  \cdots \)

\(\ln \left( {x + 1} \right) \buildrel\textstyle.\over= x - \frac{1}{2}{x^2} + \frac{1}{3}{x^3} - \frac{1}{4}{x^4} +  \cdots \)