Answer
$\Sigma_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k} (x-1)^{k}$
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=\ln (x)$
Differentiate
$f'(x)=\dfrac{1}{x}$
$f''(x)=-\dfrac{1}{x^2}$
$f'''(x)=\dfrac{2}{x^3}$
Now $f(1)=0$, $f'(1)=1$, $f''(1)=-1$, $f'''(1)=2$
Hence the Maclaurin Series for the function is
$\displaystyle (x-1)-\dfrac{1}{2}(x-1)^2+.........+ \dfrac{(-1)^{n-1}}{n!} (x-1)^{n}+....=\Sigma_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k} (x-1)^{k}$
