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.
 A: Let $u = x + 2$ so that $x = u - 2$. Then,
$$
\frac{1}{x^2} = \frac{1}{(u - 2)^2} 
= \frac{1}{(2 - u)^2}
= \frac{1}{4 \bigl(1 - \tfrac12 u \bigr)^2}
= \frac{1}{4} \frac{1}{(1 - v)^2}
$$
where $v = \tfrac12 u$ or $u = 2v$.
We can derive a nice power series for $(1 - v)^{-2}$ centered at $v = 0$ by differentiating the geometric series in $v$. Try it yourself.

 $$ \frac{1}{(1 - v)^2} = \frac{d}{dv} \frac{1}{1 - v} = \frac{d}{dv} \sum_{m=0}^\infty v^m = \sum_{m=0}^\infty \frac{d}{dv} v^m = \sum_{m=1}^\infty m\, v^{m-1} = \sum_{n = 0}^\infty (n+1)\, v^n,$$ where the sum is reindexed by $n = m - 1$ or $m = n+1$.

Now, back substitute $v \mapsto u \mapsto x$. Again, try to do this yourself before revealing the spoiler.

 $$ \frac{1}{x^2} = \frac{1}{4} \frac{1}{(1 - v)^2} = \frac{1}{4} \sum_{n=0}^\infty (n+1)\, \bigl(\tfrac12(x + 2) \bigr )^n = \sum_{n=0}^\infty \frac{n+1}{2^{n+2}} \, (x+2)^n$$

A: The standard riff here is to first notice that $1/x^2$ is ($-1$ times) the derivative of $1/x$... and that differentiation preserves power series with a given center. Then some algebra gives
$$ 1/x \;=\; 1/(x+2-2) \;=\; {-1\over 2}{1\over 1-{x+2\over 2}}
\;=\; {-1\over 2} \sum_{n\ge 0} ({x+2\over 2})^n
$$
Differentiating (and multiplying through by $-1$) gives
$${1\over x^2} \;=\; {1\over 2}\sum_{n\ge 0} {n\over 2}\left({x+2\over 2}\right)^{n-1}
$$
A: 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?
