I am given a geometric series $\sum_{n=0}^\infty (-\frac{1}{2})^n(x-3)^n$ and am asked to find the values of x for which the given geometric series converges then find the sum of the series (as a function of x) for those values of x.
Normally I would just plug in $x$ for $n$ but already having an $x$ inside makes me think that I can't do that.
I rearranged the expression so it now looks like this: $\sum_{n=1}^\infty (-\frac{x-3}{2})^{n-1}$ which resembles the formula for a geometric series: $\sum_{n=1}^\infty ar^{n-1}=\frac{a}{1-r}, |r|<1$. I know that this is important because the problem tells us that the series converges. And if $\sum_{n=1}^\infty a_n$ converges, then $a_n \rightarrow 0$. I believe that this would allow me to set $\sum_{n=1}^\infty ar^{n-1}=\frac{a}{1-r}=0$
So now $a=1$ and $r=-\frac{2}{x-3}$ but I don't know how to show that $|r|<1$ or if it even is. Since the summation starts at $x=1$, is it acceptable to plug in $1$ for $x$ where $r=-\frac{2}{x-3}$? If it isn't and/or if I have been doing this wrong the whole time, what steps would I need to take to solve for x?
Thanks for your time.