How to develop $f(x)=-\frac{1}{(x+2)^2}$ into a power series?

Question

I tried to develop the following $f(x)=-\frac{1}{(x+2)^2}$

into a power series. However, I was not able to do that. Since I tried to split it into two fractions, and it didn't work neither for me or when Trying on Wolfram, I am stuck.

So then I tried computing the power series by the following procedure:

\begin{equation} f(x)=-\frac{1}{(x+2)^2}=-\left(h(x)\right)^2\\ h(x)=\frac{1}{x+2}=\frac{1}{2}\frac{1}{1-(-\frac{x}{2})}\\ = \frac{1}{2}\sum_{n=0}^\infty\left(-\frac{x}{2}\right)^n\\ = \sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}x^n\\ f(x)=-\left(\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}x^n\right)^2 \end{equation}

But from here I am stuck.

Any hints?

Thanks

Answer 1

So you want to write series...I suppose a Maclaurin series, around $\;x=0\;$ . Then you can go as follows:

$$-\frac1{(x+2)^2}=\left(\frac1{x+2}\right)'=\frac12\left(\frac1{1+\frac x2}\right)'=\left[\frac12\sum_{n=0}^\infty(-1)^n\left(\frac x2\right)^n\right]'=\ldots$$

continue...and remember a power series can be termwise differentiated within its convergence interval...

Answer 2

Since your OP completely changed, your new question needs a new answer.

Your idea was good! Persevere:

$$f(x)=-\left(\sum_{k=0}^\infty\frac{(-1)^k}{2^{k+1}}x^k\right)^2=\sum_{n=0}^\infty a_nx^n$$ with $$a_n=-\sum_{i+j=n\;(i,j\in\mathbb N)}\frac{(-1)^i}{2^{i+1}}\frac{(-1)^j}{2^{j+1}} =-\sum_{i=0}^n\frac{(-1)^n}{2^{n+2}}=\frac{(-1)^{n+1}(n+1)}{2^{n+2}}.$$ The keyword is Cauchy product.