<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] Binomial theorem:
$\begin{align*}
  \frac{1}{x^2}
    &= (2 + (x - 2))^{-2} \\
    &= \frac{1}{4} 
              \cdot \left(
                            1 + \frac{1}{2}(x - 2)
                        \right)^{-2} \\
    &= \frac{1}{4}
              \cdot \sum_{k \ge 0}
                             \binom{-2}{k}
                                 \left(
                                    \frac{x - 2}{2}
                                \right)^k \\
    &= \sum_{k \ge 0} \frac{(-1)^k}{2^{k + 2}}
                            \binom{k + 1}{2} (x - 2)^k \\
    &= \sum_{k \ge 0}
              \frac{(-1)^k (k + 1) k}{2^{k + 3}}(x - 2)^k
\end{align*}$
Any way you get a power series that converges to your function will give the same series. What way you select depends on taste/ease/familiarity.
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