Answer
$\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x-1)^{n+1}}{(n+1)}$
Work Step by Step
Here, we have: $f(x)=\ln x \implies f(1)=0\\ f'(x)=\dfrac{1}{x} \implies f'(1)=1\\ f''(x)=-\dfrac{1}{x^2} \implies f''(1)=-1$
and so on.
The Taylor series of a function $f(x)$ at $x=c$ can be written as:
$T(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+.......\dfrac{f^n(a)}{n!}(x-a)^n$
Hence, the required Taylor series is:
$T(x)=0+\dfrac{1}{1!}(x-1)+\dfrac{-1}{2!} (x-1)^2+....=(x-1)--\dfrac{(x-1)^2}{2}+.......=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(x-1)^{n+1}}{(n+1)}$
