# Power Series Expansion

How can I find the Maclaurin series for $f(x)=e^x$/$(1-x^2)$? I have tried expanding it out but I am having trouble with the algebra of it.

One may recall that, as $x \to 0$, \begin{align} e^x&=\sum_{n=0}^{\infty}\frac{x^n}{n!}, \quad x \in \mathbb{C}, \tag1\\\\ \frac{1}{1-x^2}&=\sum_{n=0}^{\infty}x^{2n}=\frac1{2}\sum_{n=0}^{\infty}(1+(-1)^n)x^{n}, \quad |x|<1, \tag2 \end{align} then using the Cauchy product we get
$$\frac{e^x}{1-x^2}=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\frac{(1+(-1)^{n-k})}{2k!}\right)x^{n}, \quad |x|<1. \tag3$$
The series on the right hand side of $(3)$ starts as $$\frac{e^x}{1-x^2}=1+x+\frac3{2}x^2+\frac7{6}x^3+\cdots$$
• A small nitpick: (1) and (2) are not only when $x\to 0$ (the first holds for all $x\in\mathbb{C}$, the second when $\lvert x\rvert \leq 1$). May 21 '15 at 13:59
• @ClementC. It is not true that the second holds when $|x|\leq1$ :) May 21 '15 at 18:06
• Sorry. $\lvert x\rvert < 1$, of course... nitpicking away! May 21 '15 at 18:36