I am trying to get the power series for $\ln(9-4x^2)$
The first step I took was taking the derivative of $\ln(9-4x^2)$ which I got: $\frac{-8x}{9-4x^2}$
Getting this into the form of $\sum_{n=1}^\infty ar^n$ = $\frac{a}{1-r}$ is where I am having issues.
Assuming that you are interested in a power series centered at $0$, note that\begin{align}-\frac{8x}{9-4x^2}&=-\frac{8x}9\times\frac1{1-\frac49x^2}\\&=-\frac{8x}9\sum_{n=0}^\infty\left(\frac49\right)^nx^{2n}\text{ (if $|x|<\frac32$)}\\&=-2\sum_{n=0}^\infty\left(\frac49\right)^nx^{2n+1}.\end{align}Therefore, the power series that you are after is$$\log(9)-2\sum_{n=0}^\infty\frac{(4/9)^n}{2n+2}x^{2n+2}=\log(9)-\sum_{n=0}^\infty\frac{(4/9)^n}{n+1}x^{2n+2}$$
$$\ln(9-4x^2)=\ln9+\ln\left(1-\dfrac{4x^2}9\right)$$
Now use $$\ln(1-y)=-\sum_{r=1}^\infty\dfrac{y^r}r$$ for $-1\le y<1$
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