Answer
$$\dfrac{x^3}{1+x^2}; |x| \lt 1$$
Work Step by Step
The Taylor series for $\dfrac{1}{ (1-x)}$ can be defined as: $\dfrac{1}{ (1-x)}=1+x+x^2+.....x^n+....; |x| \lt 1$
Consider the given series:
$\\=x^3-x^5+x^7-x^9-....\\=x^3(1-x^2+x^4-x^6+....)\\=x^3(1-x^2+(x^2)^2-(x^2)^3+.......) \\ = x^3 (\dfrac{1}{1+x^2}) \\=\dfrac{x^3}{1+x^2}; |x| \lt 1$
