<s>[INST] Fibonacci power series and fraction decomposition In a recent problem I have established that if $|x|<\frac{1}{2}$ then 
$$f(x)=\sum_{n=1}^{\infty}a_{n}x^{n-1}=\frac{-1}{x^2+x-1}$$
where $a_{1}=a_{2}=1$ and $a_{n+1}=a_{n}+a_{n-1}$
(To work out the sum to the fraction, simply note that if $f(x)$ represents the sum then $f(x)-xf(x)-x^2f(x)=1)$)
I am now asked to find another series representation using the fraction $\frac{-1}{x^2+x-1}$
By fraction decomposition we have that 
$$\frac{-1}{x^2+x-1}=\frac{\frac{1}{\sqrt{5}}}{x-\frac{-1-\sqrt{5}}{2}}+\frac{\frac{-1}{\sqrt{5}}}{x-\frac{-1+\sqrt{5}}{2}}$$
I now thought that i could transform the expressions into the form $$\frac{1}{1-(\frac{x}{a})}=\sum_{n=0}^{\infty}\left(\frac{x}{a}\right)^n$$
which I went about as follows (starting from the decomposite version):
\begin{align*}
\frac{\frac{1}{\sqrt{5}}}{x-\frac{-1-\sqrt{5}}{2}}+\frac{\frac{-1}{\sqrt{5}}}{x-\frac{-1+\sqrt{5}}{2}}
=&\frac{1}{\sqrt{5}}\left(\frac{1}{x-\frac{-1-\sqrt{5}}{2}}-\frac{1}{x-\frac{-1+\sqrt{5}}{2}}\right) \\
=&\frac{1}{\sqrt{5}} \left(\frac{1}{\frac{2x}{-1-\sqrt{5}}-1}-\frac{1}{\frac{2x}{-1+\sqrt{5}}-1}\right) \\
=&\frac{1}{\sqrt{5}} \left(\frac{-1}{1-\frac{2x}{-1-\sqrt{5}}}+\frac{1}{1-\frac{2x}{-1+\sqrt{5}}}\right)\\
=& \frac{1}{\sqrt{5}}\left[-\sum_{n=0}^{\infty}\left(\frac{2}{-1-\sqrt{5}}\right)^n x^n+\sum_{n=0}^{\infty}\left(\frac{2}{-1+\sqrt{5}}\right) x^n\right] \\
=& \frac{1}{\sqrt{5}}\sum_{n=0}^{\infty}\left[\left(\frac{2}{-1+\sqrt{5}}\right)^n-\left(\frac{2}{-1-\sqrt{5}}\right)^n\right]x^n
\end{align*}
However wolfram says that the last sum evaluates to $\frac{-x}{x^2+x-1}$ which is one x to much. I would appreciate any help on where I went wrong!
 [/INST] On the second line of your equalities, you multiplied denominators but not numerator by $$\frac2{-1\pm\sqrt5}.$$
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