Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$. 
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore, $\frac{d}{du} (-\frac{1}{u-2})= \frac{1}{x^2}$ and
$$\frac{d}{du} (-\frac{1}{u-2})= \frac{d}{du} (-\frac{1}{-2(1-\frac{u}{2})})=\frac{d}{du}(\frac{1}{2} \frac{1}{1-\frac{u}{2}})=\frac{d}{du} \Bigg( \frac{1}{2} \sum_{n=0}^\infty \bigg(\frac{u}{2}\bigg)^n\Bigg)$$
$$= \frac{d}{du} \Bigg(\sum_{n=0}^\infty \frac{u^n}{2^{n+1}}\Bigg)= \frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=$$
$$\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
From this we can conclude that 
$$f(x)=\frac{1}{x^2}=\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
Is this solution correct?

Why not a Taylor series expansion at $x=-2$? That would be
$$f(x) = \sum_{n\geq 0} \frac{f^{(n)}(-2)}{n!}(x+2)^n, $$
with radius of convergence of $2$. You may calculate the $n$-th derivative of $f(x)=1/x^2$ to find
$$\frac{d^nf}{dx^n}(x)= \frac{(-2)(-3) \cdots (-2-n+1)}{x^{n+2}} = (-1)^n \frac{(n+1)!}{x^{n+2}} $$
and so
$$\frac{1}{x^2} = \sum_{n\geq 0} \frac{(n+1)}{2^{n+2}}(x+2)^n, \qquad x\in(-4,0) .$$
