Answer
$\sum_{n = 0}^{\infty}\frac{(-1)^n x^{4n+2}}{2^{4n+4}}$
Interval: $(-2, 2)$
Work Step by Step
f(x) = $\frac{x^2}{x^4 + 16}$ = $\frac{x^2}{2^4} (\frac{1}{1+(\frac{x}{2})^4}) $
= $\frac{x^2}{2^4} (\frac{1}{1-(-1)(\frac{x}{2})^4})$
= $\frac{x^2}{2^4} \sum_{n = 0}^{\infty} \frac{(-1)^n x^(4n)}{2^(4n)}$
= $\sum_{n = 0}^{\infty} \frac{(-1)^n x^{4n+2}}{2^{4n+4}}$
Interval:
The series converges when |$\frac{x^4}{2^4}$|<1, hence when |$x^4$| < $2^4$, so $R = 2$, Interval = $(-2, 2)$
