<s>[INST] Taylor Series of $ \cfrac{x}{1+x} $ at $ x = -2 $ I am trying to find the taylor series for the following:
$$ \cfrac{x}{1+x} \ \text{ at } \ x = -2 $$
The following are my steps:
$$ \cfrac{x}{1+x} = 1 - \cfrac{1}{1+x} \ \text{ (by long division) } $$
Taylor series of $ \cfrac{1}{1+x} $ at $ x = 0, $
$$ \begin{align} \cfrac{1}{1+x} = \displaystyle\sum_{n=0}^{\infty}{(-1)^n(x)^n}\end{align} $$
Taylor series of $ \cfrac{1}{1+x} $ at $ x = -2, $
$$ \begin{align} \cfrac{1}{1+(x+2)} = \displaystyle\sum_{n=0}^{\infty}{(-1)^n(x+2)^n}\end{align} $$
Taylor series of $ \cfrac{x}{1+x} $ at $ x = -2, $
$$ \begin{align}
\cfrac{x}{1+x}
&= 1 - \cfrac{1}{1+x} \\
&= 1 - \displaystyle\sum_{n=0}^{\infty}{(-1)^n(x+2)^n} \\ 
&= 1 - \left(\cfrac{1}{1+(-2)} + \displaystyle\sum_{n=1}^{\infty}{(-1)^n(x+2)^n}\right) \\ 
&= 2 - \displaystyle\sum_{n=1}^{\infty}{(-1)^n(x+2)^n} 
\end{align}
$$
Somehow, the answer is: 
$ 2 + \displaystyle\sum_{n=1}^{\infty}{(x+2)^n}  $
I know that my last block of working seems wrong. But I just cannot think of anything else. Could someone please advise me?
 [/INST] Hint. Note that
$$\frac{x}{1+x}=\frac{x}{-1+x+2}=\frac{-(x+2)+2}{1-(x+2)}=1+\frac{1}{1-(x+2)}.$$
Now use the result: for $|z|<1$ then
$\displaystyle \cfrac{1}{1-z} = \displaystyle\sum_{n=0}^{\infty}{ z^n}$.
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