Answer
\begin{aligned}
\ln (1+x) =x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+(-1)^{n-1} \frac{x^{n}}{n}
\end{aligned}
Work Step by Step
Since
\begin{aligned}
f(x) &=\ln (x+1) & & f(0)=0 \\
f^{\prime}(x) &=\frac{1}{1+x} & & f^{\prime}(0) =1 \\
f^{\prime \prime}(x) &=\frac{-1}{(1+x)^{2}} & & f^{\prime \prime}(0) =-1 \\
f^{\prime \prime \prime}(x) &=\frac{2}{(1+x)^{3}} & & f^{\prime \prime \prime}(0) =2 \\
f^{(4)}(x) &=\frac{-6}{(1+x)^{4}} & & f^{(4)}(0) =-6
\end{aligned}
Then
\begin{aligned}
\ln (1+x) &=f(0)+\frac{f^{\prime}(0)}{1 !} x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\ldots \ldots+\frac{f^{(n)}(0)}{n !} x^{n} \\
&=\frac{1}{1 !} x-\frac{1}{2 !} x^{2}+\frac{2}{3 !} x^{3}-\frac{6}{4 !} x^{4} \ldots \\
&=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+(-1)^{n-1} \frac{x^{n}}{n}
\end{aligned}
