Answer
$\ln (1+x) =\Sigma_{n=1}^{\infty} (-1)^{n-1} \dfrac{x^n}{n}$
Work Step by Step
We know that the Maclaurain series for $\ln x=\Sigma_{n=1}^{\infty} (-1)^{n-1} \dfrac{(x-1)^n}{n}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+.....$
Replace $x$ with $1+x$ in the above series.
$\ln (1+x) =\Sigma_{n=1}^{\infty} (-1)^{n-1} \dfrac{(1+x-1)^n}{n}$
Thus, our required series is:
$\ln (1+x) =\Sigma_{n=1}^{\infty} (-1)^{n-1} \dfrac{x^n}{n}$
