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Expand $\frac{x}{4-x^2}$ as a power series around $x_0 = 0$ and compute its radius of convergence.

$\frac{x}{4-x^2} = x \frac1{4-x^2} = \frac{x}4 \frac{1}{1-\frac{x^2}{4}} = \frac{x}4\sum_{n=0}^\infty (\frac{x^2}{4})^n \text{(assuming that} \frac{x^2}{4} <1) = \sum_{n=0}^\infty \frac1{4^{n+1}} x^{2n+1}$.

I am stuck here. How can I proceed from here to get the power series : $\sum_{n=0}^\infty a_n (x - 0)^n$?

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    $\begingroup$ You are almost done. The $a_n$ will be piecewise defined, depending on if $n$ is even or odd. $\endgroup$ – Clayton Dec 18 '19 at 2:31
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Use

$$\frac1{1-t} = 1+ t+t^2+t^3+\>...$$

to expand

$$\frac{x}{4-x^2}=\frac x4\cdot\frac1{1-(\frac x2)^2} =\frac x4\cdot \left( 1 + (\frac x2)^2 + (\frac x2)^4+(\frac x2)^6+\>... \right)$$ $$=\frac x{2^2} + \frac {x^3}{2^4} + \frac {x^5}{2^6}+\frac {x^7}{2^8}+\>... = \sum_{k=0}^\infty \frac{[1-(-1)^k]}{2^{k+2}}x^k $$

Its radius of convergence is $(\frac x2)^2 < 1$, or $|x|<2$.

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Hint: $\dfrac{x}{4-x^2} = \dfrac{1}{4}\left(\dfrac{1}{1-\frac{x}{2}} - \dfrac{1}{1+\frac{x}{2}}\right)$. Note that this way of expanding is essentially the same as the one you have. But going this way helps you not only to keep track of the coefficients $a_n$ better, but to write a formula for the $a_n$ easier as well.

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