Question about taylor series$ -((2/x)/2!)$

Question

right now I'm learning about taylor series in uni and I'm getting kinda confused. The question is : find the polynomials of order $0,1,2,3$ generated by $f$ at $a$. $f(x) = 2 \ln (x)$, $a = 1$. I can do out the question just fine, however when you get to the point of having $-((2/x^2)/2!)$ , why does this become $-(2/2!)$?

Answer 1

Hint

Since $a=1$, we consider the function $$g (x)=f (x+1)=2\ln (x+1) $$

$$=2 (x-\frac {x^2}{2}+\frac {x^3}{3})+x^3\epsilon (x)$$

thus

$$f (x)=g (x-1)$$ $$=2(x-1)-(x-1)^2+\frac {2(x-1)^3}{3}+(x-1)^3\epsilon (x) $$