# Exercise 2.6.6 Introduction to Real Analysis by Jiri Lebl

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 : $$\sum_{n=0}^\infty a_n (x - 0)^n$$?

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

$$\frac1{1-t} = 1+ t+t^2+t^3+\>...$$
$$\frac{x}{4-x^2}=\frac x4\cdot\frac1{1-(\frac x2)^2} =\frac x4\cdot \left( 1 + (\frac x2)^2 + (\frac x2)^4+(\frac x2)^6+\>... \right)$$ $$=\frac x{2^2} + \frac {x^3}{2^4} + \frac {x^5}{2^6}+\frac {x^7}{2^8}+\>... = \sum_{k=0}^\infty \frac{[1-(-1)^k]}{2^{k+2}}x^k$$
Its radius of convergence is $$(\frac x2)^2 < 1$$, or $$|x|<2$$.
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.