## Calculus 8th Edition

$f(x)=\sum_{n=0}^{\infty} c_{n}x^{n}=\frac{(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3})}{1-x^{4}}$ $R=1$, with interval of convergence is $(-1,1)$
$\sum_{n=0}^{\infty} c_{n}x^{n}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+....+$ Separate this series into 4 chunks. $\sum_{n=0}^{\infty} c_{n}x^{n}=(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3})+x^{4}(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3})+....$ we get $f(x)=\sum_{n=0}^{\infty} c_{n}x^{n}=\frac{(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3})}{1-x^{4}}$ $R=1$, with interval of convergence is $(-1,1)$