A question regarding representation of a function as a power series

I'm trying to help my brother with a calculus problem related to the representation of a function as a power series. The task is to find what power series is represented by the following function, and what is its interval of convergence:

$$f(x)= \frac{4}{4 + x^2} \text{, where the center of the series is }c=0 \text{.}$$

$\textbf{My attempt at a solution:}$ I write $\frac{4}{4+ x^2}= \frac{1}{1+ x^2/4}$. Now I know that $\sum_{n=0}^\infty ar^n= \frac{a}{1-r}$, so in our case I believe we have $a=1, r=-\frac{x^2}{4}$. From this I conclude that

$$\frac{4}{4+ x^2}= \sum_{n=0}^\infty \left(-\frac{x^2}{4}\right)^n \text{.}$$

This series converges for $|-\frac{x^2}{4}| < 1$, implying that $|x^2|<4$, and thus $|x|<2$, right? If this is correct, then that means the radius of convergence for this series is $R= 2$. I checked that this series diverges for $x= -2,2$. Therefore, we get that the interval of convergence for this series is the open interval $(-2,2)$. Was this the correct approach to be using? Please let me know if there are any mistakes, thanks.

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Looks good to me. – Christopher A. Wong Mar 30 '13 at 0:11
Yes, you came up with the right answer. There's an easier way to go about finding the solution, which depends on the knowledge that the power series of $\frac{1}{1-x}$ is $\sum_{n=0}^\infty x^n$. You may then use the replacement $x\mapsto -\frac{x^2}{4}$ and achieve the same result you did. – Ian Coley Mar 30 '13 at 0:13
+1 Looks fine to me. – DonAntonio Mar 30 '13 at 0:15
@Christopher Certainly effectively, but his logic appeared to use the definition of the sum of a geometric series rather than the power series of the function. – Ian Coley Mar 30 '13 at 0:30
Doesn't look quite finished to me. – Mark McClure Mar 30 '13 at 17:17

As the comments have indicated, what you've done is correct, but I wouldn't agree that it's quite complete. You still need to express it as a power series, i.e. as a sum of coefficients times powers of $x$, not as powers of $-x^2/4$. Thus, the final step would be

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

To illustrate why this step is important (beyond merely satisfying the definition of power series), supposed you want to go to some further application and, say, integrate the result. Integration is easy, once you've expressed the result as a sum of powers, but it's not so easy in the form you've expressed it since there's no "chain rule" for integration. That's rather an important point of power series - they provide us with an alternative representation of the function and certain operations are easy given this alternate representation.

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