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).
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.