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This question has been asked already on this site, but I found the solution confusing relative to my understanding.

Use differentiation to find a power series representation for: $f(x)=\frac{1}{(8+x)^2}$

My process is:

I can represent it as $\frac{d}{dx}\left(\frac{-1}{8+x}\right)$

now $$\frac{-1}{8+x}= \frac{-1}{8}\frac{1}{\left(1+\left(\frac{x}{8}\right)\right)}$$

Representing the original function as a power series:

$$\frac{d}{dx}\left(-\sum\limits_{n=0}^{\infty}\frac{(-1)^nx^n}{8^{n+1}}\right)$$

My question is, how can we treat the negative on the outside? Are we allowed to factor it in. Because if I do, I don't get the right answer (I put it in to wolfram-alpha). Leaving it outside also gives me a wrong answer (the sum should not have a negative sign behind it).

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    $\begingroup$ you could change $-(-1)^n$ to $(-1)^{n\pm1}$, if that's what you're asking $\endgroup$ – J. W. Tanner Apr 10 '19 at 3:35
  • $\begingroup$ It would be easier to answer this question if you would tell us what answer you get and what answer Wolfram gets. Without that, we can only guess as to what you might have done. $\endgroup$ – Gerry Myerson Apr 10 '19 at 3:42
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I managed to get the answer. I thought that $-(-1)^n$ was $1^{n+1}$, or just $1$, but it's simply $(-1)^{n+1}$, which would yield the answer.

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