<s>[INST] Representing $\frac{1}{x^2}$ in powers of $(x+2)$ I was asked to represent $f(x)=\frac{1}{x^2}$ in powers of $(x+2)$ using the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I am able to represent $\frac{1}{x^2}$ as a power series, but I am struggling withdoing it in powers of $(x+2)$. This is what I attempted.
$I.$ Firstly, I used the simple expansion
$$\ln(x+1) = \sum_{n=1}^\infty (-1)^n\frac{x^n}{n}$$
which comes from the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I then noticed that
$$\ln(x) = \ln(1+(x-1))=\sum_{n=1}^\infty (-1)^n\frac{(x-1)^n}{n}$$
$II$. $-f(x)=-\frac{1}{x^2}$ is the second derivative of $\ln(x)$, so that
$$\frac{d}{dx}\sum_{n=1}^\infty (-1)^n\frac{(x-1)^n}{n}= \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-1}}{n^2}$$
and
$$\frac{d}{dx} \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-1}}{n^2} = \sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-2}}{n^2(n-1)}=-\frac{1}{x^2}$$
from what plainly follows that $$\frac{1}{x^2} = -\sum_{n=1}^\infty(-1)^n \frac{(x-1)^{n-2}}{n^2(n-1)}$$
$III.$ Of course, this expansion is not in powers of $(x+2$). What I pressumed would be the logical thing to do is to repeat these steps but, instead of the using $\ln(1 + (x-1))$, using the equivalent
$$\ln(-2 + (x+2))=\ln((-2)(1+\frac{x+2}{(-2)})) = \ln(-2)+\ln(1+\frac{x+2}{(-2)})$$
I expected to be able to use this because the second term is of the form $\ln(1+ u)$ where $u$ is some function of $x$, and we know the expansion of such expression. However, $\ln(-2)$ is nonsense, since $\ln(x)$ is defined only for $\mathbb{R}^+$.
Is there an alternative, better way to do this or am I missing something? Thanks.
 [/INST] There is a better way.
$\frac 1  {x^{2}}=\frac 1  {((x+2)-2)^{2}}=\frac  1 4 \frac  1 {(1-\frac {x+2} 2)^{2} } =\frac 1  2\frac  d {dx} \frac  1{1-\frac {x+2} 2}=\frac 1 2\frac d {dx} [\sum\limits_{n=0}^{\infty} (\frac {x+2} 2 )^{n}]$. Can you continue?
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