Power Series and Functions — Question 10

Question 10 – Solution

Step 1: Expand the reciprocal.

Use the geometric ratio $r=-x^3$: $$\frac1{1+x^3}=\frac1{1-(-x^3)}=\sum_{n=0}^{\infty}(-1)^n x^{3n},\qquad |x^3|<1.$$

Step 2: Multiply by $x^2$.

$$\boxed{\frac{x^2}{1+x^3}=\sum_{n=0}^{\infty}(-1)^n x^{3n+2}}.$$ The exponents are $2,5,8,\ldots$ because multiplying by each additional ratio contributes three powers of $x$.

Step 3: Determine convergence.

The interior condition is $|x|<1$, so $R=1$. At $x=1$, terms are $(-1)^n$; at $x=-1$, every term is $1$. Neither tends to zero.

Conclusion.

The interval is $\boxed{(-1,1)}$.