Find taylor series for $x/1+x$, at $x= -2$ using differentiation Attempt:
$\\f^ 1(x) = \frac{1}{(1+x)^2}$ ,
$\\f^ 2(x) = \frac{-2}{(1+x)^3}$ ,
$\\f^ 3 (x) = \frac{6}{(1+x)^4}$
$\\f^ n (x) = \frac{(-1)^{n+1}(n!)}{(1+x)^{n+1}}$,
$\\f^ n (-2) = \frac{(-1)^{n+1}(n!)}{(-1)^{n+1}}$
Using the taylor series formula,
$\frac {x}{1+x} = \sum_{n=0}^{\infty} \frac{f^n(-2)}{n!} (x+2)^n = \sum_{n=0}^{\infty} \frac{n!}{n!}(x+2)^n
= \sum_{n=0}^{\infty} (x+2)^n $
But the solution used geometric series and $\frac {x}{1+x} = 1 + \sum_{n=0}^{\infty} (x+2)^n$.
I am unsure as to what is wrong with my solution, is there a  way to get this  correct answer using taylor series formula directly instead of using geometric series?

Hint
Make life simpler letting $x=y-2$. So
$$\frac x{1+x}=\frac{y-2}{y-1}=1+\frac{1}{1-y}$$
The derivatives wrt $y$ are very simple and, when finished, make $y= {x+2}$
