calculus - Manipulating a Power Series to change convergence at endpoints

The problem is to find a power series that satisfies convergence on the interval $[-1,3), (-1,3), (-1,3],$ and $[-1,3]$. I have worked out a general form equation that satisfies the first two, but I am having issues trying to manipulate the formula for parts $3$ and $4.$

General Eq.

$$\sum_{n=1}^\infty\frac{(x-1)^n}{2^n}$$ for $(-1,3)$
and $$\sum_{n=1}^\infty\frac{(x-1)^n}{n2^n}$$ for $-1,3)$

No matter what I try It seems like the numerator will always be

$$2^n$$ or the bounds of convergence change. If it isn't possible would my explanation be the reason why?

Answer

For the $(-1,3]$, the idea used for $[-1,3)$ will work. Just multiply by $(-1)^n$.

For $[-1,3]$, try $\sum \frac{(x-1)^n}{n^2 2^n}$.