<s>[INST] 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?
 [/INST] Your answer is extremely close to the correct derivation. The error occurs when you write

$$\frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=\sum_{\color{red}{n=0}}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$

where the last equality should have an index starting from $n=1$. The correct derivation is
\begin{align}\frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)&=\sum_{\color{blue}{n=1}}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}\\&=\sum_{\color{blue}{n=0}}^\infty \frac{\color{blue}{(n+1)}}{2^{\color{blue}{(n+1)+1}}} (x+2)^{\color{blue}{(n+1)-1}}\\&=\sum_{n=0}^\infty \frac{(n+1)}{2^{n+2}} (x+2)^{n}\end{align}
You could also find the power series through giobrach's answer, which might be the most straightforward technique.
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