What are the conclusions on convergence and divergence based on $\sum_{n=0}^\infty a_n * x^n$?

Question

Take the sum: $$\sum_{n=0}^\infty a_n * x^n$$ and assume convergence at $x=-3$ and divergence at $x=9$. What can you conclude about $$\sum_{n=0}^\infty (-1)^n * a_n * 4^n$$ and $$\sum_{n=0}^\infty a_n * 8^n$$ I assume this has something to do with the radius, $R$, and interval of convergence. However, I am having difficulty determine exactly what those are.

My thought was if you apply the ratio test to $$\sum_{n=0}^\infty a_n * x^n$$ you eventually get $$\lim\limits_{x \to \infty} \frac{a_{n+1}}{a_n} * |x|$$ If $$\lim\limits_{x \to \infty} \frac{a_{n+1}}{a_n} \ne \infty$$ and $$\lim\limits_{x \to \infty} \frac{a_{n+1}}{a_n} \ne 0$$ then $$\lim\limits_{x \to \infty} \frac{a_{n+1}}{a_n} = c$$ and for a number $$(c|x|) < 1$$ the sum converges. Since we are given numbers of convergence and divergence, can we assume this is what happens? Since we know the sum diverges at $x=9$, would $c=\frac{1}{9}$? Thus, making the interval of convergence, $-9<x<9$ or $(-9,9)$?

Assuming the above, the series $$\sum_{n=0}^\infty (-1)^n * a_n * 4^n$$ and $$\sum_{n=0}^\infty a_n * 8^n$$ would converge since $4$ and $8$ $\in (-9,9)$. I don't believe this is what happens though.

Answer 1

You have a power series. From the first condition we can conclude that the radius of convergence $R$ is $3 \le R \le 9$. Hence, in general we cannot say anything about the series at $x = -4$ and $x = 8$. For example, if $R = 3.5$ then both series $\sum (-4)^n a_n$ and $\sum 8^n a_n$ diverge, if $R = 5$ then $\sum (-4)^n a_n$ converges and the second one diverges and etc.

Answer 2

You can't conclude much; you can design a series which converges at $x=-3$, diverges at $x = 9$ and (1) converges at both $x = -4$ and $x = 8$, (2) converges at $x = -4$ but diverges at $x = 8$ or (3) diverges at both $x=-4$ and $x=8$. Indeed, consider $\alpha> 0$. Then $$\frac{1}{1 - \alpha x} = \sum^\infty_{n=0} \alpha^n x^n \,\,\,\, \text{ for } \,\,\,\, \lvert x\rvert < \frac{1}{\alpha}.$$ We need $1/9 < \alpha < 1/3$ to give convergence at $x = -3$ and divergence at $x = 9$; we'll fix $\alpha$ in this range so satisfy the first two conditions. If $\alpha = 1/3.5$, we get divergence at $x = -4$ and at $x =8$. If $\alpha = 1/5$, we get convergence at $x = -4$ but divergence at $x = 8$. If $\alpha = 1/8.5$, we get convergence at both $x = -4$ and $x = 8$.

We do know that if the series converges at $x = 8$ then it must also converge at $x = -4$ since the range of convergence must be symmetric about $x = 0$.