So I know to determine an interval for a power series, we would use things like the ratio test and then some of the other tests to determine the endpoints but how would you go about doing the reverse?
If we had any such interval like (a,b], [a,b], [a,b), or (a,b), how would you go about creating a power series that would have such an interval?
An observation I was thinking of, since the general expression for a power series contains $(x-c)$ where the series is centered around $c$, we would have it so that $c$ = $(a+b)/2$
Is there a systematic way of doing such a thing?
Absolutely, and you've got a good start on it. Start with a power series with radius of convergence $1$ about the origin. Two questions:
1) How to get get a power series of radius of convergence $c$ from this?
2) How do you move the center of the interval to where you want it?
Hint: Neither of these is hard.
EDIT
Example:
Suppose we choose $\sum_{n=0}^{\infty}{x^n}$ as a series with radius of convergence $1$. How can we change it to get a series of radius of convergence $2$? All we have to do is change $x$ to $\frac{x}{2}.$ The radius of convergence of $$\sum_{n=0}^{\infty}{\left(\frac{x}{2}\right)^n}=\sum_{n=0}^{\infty}{2^{-n}x^n}$$ is $2$. The series converges when $$\left|\frac{x}{2}\right|<1\iff |x|<2.$$
Now suppose that we want a series that converges on the interval $(-1,3).$ As you noted we want a radius of convergence of $2$ and the interval should be centered at $x=1$. We just constructed a series with radius of convergence $2$, but it is centered at $0$. To move it to $1$ we just make a Taylor series with the same coefficients, but centered at $1$. $$\sum_{n=0}^{\infty}{2^{-n}(x-1)^n}$$ This converges when $$|x-1|<2\iff -2<x-1<2\iff -1<x<3$$