Power Series and Functions — Question 9

Question 9 – Solution

Step 1: Put the function in geometric form.

$$\frac1{2-x}=\frac12\frac1{1-x/2}.$$ Using the geometric identity with $r=x/2$ gives $$\boxed{\frac1{2-x}=\frac12\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^n=\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}}}.$$

Step 2: Translate the restriction.

The condition $|x/2|<1$ becomes $|x|<2$, so $R=2$.

Step 3: Test endpoints.

At $x=2$, each term equals $1/2$. At $x=-2$, the terms alternate between $\pm1/2$. Neither tends to zero, so both series diverge.

Conclusion.

The interval is $\boxed{(-2,2)}$.