Find the Taylor series of $\frac{1}{x+1} $ at $x=2$ This is what I did:
$\begin{align*}
f(x)&=&(x+1)^{-1}\\
f'(x)&=&-(x+1)^{-2}\\
f''(x)&=&2(x+1)^{-3}\\
f'''(x)&=&-6(x+1)^{-4}\\
f''''(x)&=&24(x+1)^{-5}\\
&\vdots\\
f^{(n)}(x)&=&(-1)^nn!(x+1)^{-(n+1)}
\end{align*}$
Then I substituted $x=2$:
$f^{(n)}(2)=(-1)^nn!(2+1)^{-(n+1)}=(-1)^nn!(3)^{-(n+1)} $
Then I put it all together.
\begin{align*}
\frac{1}{x+1}&= &\sum_{n=0}^{\infty}\frac{f^{(n)}(2)}{n!}(x-2)\\
&=&\sum_{n=0}^{\infty}\frac{(-1)^nn!(3)^{-(n+1)}}{n!}(x-2)\\
&=&\sum_{n=0}^{\infty}(-1)^n(3)^{-(n+1)}(x-2)
\end{align*}
I am not sure if this is correct.

This is the approach mentioned by Andre:
$$\frac{1}{x+1}=\frac{1}{3+(x-2)}=\frac{1}{3}\frac{1}{1+\frac{x-2}{3}}=\frac{1}{3}\left(1-\frac{x-2}{3}+\left(\frac{x-2}{3}\right)^2-\ldots\right)=$$
$$=\frac{1}{3}\sum_{n=0}^\infty(-1)^n\left(\frac{x-2}{3}\right)^n$$
The above being true whenever 
$$\left|\frac{x-2}{3}\right|<1\Longleftrightarrow |x-2|<3\Longleftrightarrow -1<x<5$$
which is a reasonable assumption as we want $\,x\,$ to be "close" to $\,2\,$ ("close" may be in the eyes of the beholder...)
