# nick's question at Yahoo! Answers (Maclaurin series)

MHB Math Helper

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Hello nick,

If $|t|<1$ we know that $\dfrac{1}{1-t}=\displaystyle\sum_{n=0}^{\infty}t^n$. Then, using the Algebra of series: $$f(x)=x\cdot\dfrac{1}{1-x^4}=x\sum_{n=0}^{\infty}(x^4)^n=\sum_{n=0}^{ \infty}x^{4n+1}\quad (|x|<1)$$ So, necessarily the Maclaurin series for $f(x)$ is $\displaystyle\sum_{n=0}^{\infty}x^{4n+1}.$