Taylor Series of $f(x) =\frac{1}{x^2}$ I've found the Taylor Series for $f(x)=\frac{1}{x^2}$ centered at $a=-1$.
$f(-1)=1$, $f'(-1)=2$, $f"(-1)=6$, $f'''(-1)=24$, $f^4(-1)=120$
I used this formula to get each the first coefficients of the terms of the series: $c_n=\frac{f^n(a)}{n!}$
So I got the expansion:
$$f(x)=1+2(x+1)+3(x+1)^2+4(x+1)^3+5(x+1)^4+...$$
What is difficult for me is expressing this in summation notation. Is this right?
$$\sum\limits_{n=0}^\infty (n+1)(x+1)^n$$

For $x$ such that $|x+1|<1$ 
$$f(x)=\frac{-1}{x}=\frac{1}{1-(x+1)}$$
$$=1+(x+1)+(x+1)^2+....$$
$$=\sum_{k=0}^\infty (x+1)^k$$
As a sum of a power series, 
$f $ is differentiable at $ (-2,0)$ and
$$f'(x)=\frac{1}{x^2}=\sum_{k=1}^\infty k(x+1)^{k-1}$$
$$=\sum_{k=0}^\infty(k+1)(x+1)^k$$
