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