I am looking to create a power series that converge for $[8,11)$ I tried first choosing $|x-9.5|< \frac{3}{2}$ as the radius of convergence, and made it to the following power series: $$\sum_{n=1}^\infty \frac{{2^n}{(x-9.5)}^n}{3^n}$$
how can i adjust the power series in such manner that i can "control" the convergence on the ends of $8$ and $11$?
I am looking for more of general guidance regarding these sort of math problems, because the other way around is usually easier, but with these questions i tend to struggle more..
With these sorts of problems, you benefit from recalling other problems that you've solved (perhaps from the opposite direction).
If you know any example of a series which converges for $[-1, 1)$, such as $$ \sum_{n \geq 1} \frac{x^n}{n},$$ then you can adjust the width of the interval by scaling and the center of the interval by shifting. For instance, you can get $[-2, 2)$ (scaling by $2$) by dividing $x$ by $2$, or rather $$ \sum_{n \geq 1} \frac{(x/2)^n}{n}.$$ You can get $[0, 2)$ by recentering around $1$, getting $$ \sum_{n \geq 1} \frac{(x-1)^n}{n}.$$
Through this sort of thinking, you can answer all questions of this sort. Note how reminscent this is of the classic analysis of changes to a function $f(x)$: you can scale the domain by considering $f(rx)$ for some real number $r$, and you can shift the domain by considering $f(x-c)$ for some number $c$. This is really the exact same analysis, but in a different setting.