# Given $\lim_{n\to\infty} |a_{n}|^{1/n} = \alpha$, prove the following properties.

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.

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.
• How is $(1)$ an immediate application of the Root Test? – joseph Dec 10 '18 at 5:47
• why is it true that $|a_{n}|^{1/n} < \epsilon$ for sufficiently large $n$? – joseph Dec 10 '18 at 6:50
• @joseph Because $\alpha =0$ in 2). – Kavi Rama Murthy Dec 10 '18 at 7:15