Proof radius of convergence for zero and infinite (power series)

Question

I deleted my last question because there was a huge mistake inside.

Given: $R$ is the radius of convergence of $\sum_{n=0}^{\infty} a_{n}x^{n}$, also suppose that $\lim_{n\rightarrow \infty} \left | \frac{a_{n+1}}{a_{n}} \right | = a$.

Then:

1. if $a = \infty$, $R = 0$

2. if $a = 0$, $R = \infty$

I would use ratio test but there isn't even a sequence given, I just got $a_{n}$ and that's it. Can't really do much if I use ratio test on it. Actually, it has already been done, no?:

$\lim_{n\rightarrow \infty} \left | \frac{a_{n+1}}{a_{n}} \right | = a$

Maybe it would be possible if I create / assume a sequence myself and then use ratio test on it?

But what I'm pretty sure is if $a = 0$ then the sequence $\left | \frac{a_{n+1}}{a_{n}} \right |$ will converge to $0$.

For $a = \infty$, the sequence will not converge to zero.

So all I have to do now is proof this? (The two sentences above this one.)

Answer 1

Why wouldn't we be able to use the ratio test?

Let $b_n=a_nx^n$

The ratio test states if we have the series $\sum_{n=0}^\infty b_n$ and we define $L=\lim_{n\rightarrow \infty} \left|\dfrac{b_{n+1}}{b_n}\right|$, then $L$ converges absolutely if $L<1$, $L$ diverges if $L>1$, and $L$ is inconclusive if $L=1$.

$$L=\lim_{n\rightarrow \infty}\left|\dfrac{a_{n+1}x^{n+1}}{a_nx^n}\right|=a\cdot\lim_{n\rightarrow\infty}|x|=a|x|$$

Here, we can treat $|x|=c\in\mathbb{R}$ as a constant as we observe $L$ when $a$"$=$"$\infty$ and $a=0$.

If $a$"$=$"$\infty$, then $a|x|=ac\rightarrow\infty$ and because $c$ is a constant, the series cannot converge, with the exception of $c=0$. Thus, the radius of convergence is $0$.

If $a=0$, then $a|x|=ac=0$ and because $c$ is a constant, the series converges for all values of $c$. Thus, the radius of convergence is $\infty$.