Power Series and Functions — Question 7

Question 7 – Solution

Step 1: Identify the ratio.

Apply $1/(1-r)=\sum r^n$ with $r=x^2$: $$\boxed{\frac1{1-x^2}=\sum_{n=0}^{\infty}(x^2)^n=\sum_{n=0}^{\infty}x^{2n}}.$$ Only even powers occur because each power of the ratio $x^2$ adds two to the exponent.

Step 2: Translate the geometric restriction.

$$|x^2|<1\iff |x|<1.$$ Hence $R=1$.

Step 3: Test the endpoints.

At both $x=1$ and $x=-1$, every term $x^{2n}$ equals $1$. Thus the series diverges at both endpoints.

Conclusion.

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