Answer
$\Sigma_{k=0}^n \dfrac{(-1)^{k+1}}{k}(x-1)^k$
Work Step by Step
The nth Taylor polynomial for a function $f$ is
$$p_n(x)= f(x_0) + f'(x_0)(x-x_0) +\dfrac{f''(x_0)}{2!} (x-x_0)^2 +.......+\dfrac{f^n(x_0)}{n!} (x-x_0)^n$$
$f(x)=\ln x$
Differentiate w.r.t x
$f'(x)=\dfrac{1}{x}$
$f''(x)=-\dfrac{1}{x^2}$
$f'''(x)=\dfrac{2}{x^3}$
$f''''(x)=\dfrac{-6}{x^4}$
Now $f(1)=0$, $f'(1)=1$, $f''(1)=-1$, $f'''(1)=2$, $f''''(1)=-6$
Hence the nth Taylor polynomial for a function is:
$\displaystyle (x-1)-\dfrac{1}{2}(x-1)^2+ .......+\dfrac{(-1)^{n+1}}{n}(x-1)^n=\Sigma_{k=0}^n \dfrac{(-1)^{k+1}}{k}(x-1)^k$
