Given $\lim_{n\to\infty} |a_{n}|^{1/n} = \alpha$, prove:
1) If $\alpha > 0$, show $\sum_{n = 0}^{\infty} a_{n}x^{n}$ converges if $|x| < 1/\alpha$ and diverges if $|x| > 1/\alpha$
2) If $\alpha = 0$, show that $\sum_{n = 0}^{\infty} a_{n} x^{n}$ converges for all $x \neq 0$.
My try:
1) Suppose $|x| < 1/\alpha$. Then, since $\lim_{n\to\infty} |a_{n}|^{1/n} = \alpha$, we have
$$\lim_{n\to\infty} |a_{n}x^{n}|^{1/n} = \lim_{n\to\infty} |a_{n}|^{1/n}|x| = |x|\lim_{n\to\infty}|a_{n}|^{1/n} < \frac{1}{\alpha} \cdot \alpha = 1.$$
I don't know how to show convergence from this, though. I also can't show that it diverges if it's greater than $1/\alpha$.
2) For $\alpha = 0$, we have
$$\lim_{n\to\infty} |a_{n}x^{n}|^{1/n} = |x|\lim_{n\to\infty}|a_{n}|^{1/n} = |x| \cdot 0 = 0.$$
Again, I don't know if this is in the right direction.
Any help is appreciated.
1) is an immediate application of root test.
For 2) let $0 <\epsilon <\frac 1 {|x|}$. Then $|a_n|^{1/n} <\epsilon$ for $n$ sufficiently large. This gives $|a_nx^{n}| <|\epsilon x|^{n}$. Since $|\epsilon x|<1$ the geometric series $\sum |\epsilon x|^{n}$ converges. Hence the given series also converges.