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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.

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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.

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  • $\begingroup$ How is $(1)$ an immediate application of the Root Test? $\endgroup$ – joseph Dec 10 '18 at 5:47
  • $\begingroup$ See en.wikipedia.org/wiki/Root_test $\endgroup$ – Kavi Rama Murthy Dec 10 '18 at 5:48
  • $\begingroup$ why is it true that $|a_{n}|^{1/n} < \epsilon$ for sufficiently large $n$? $\endgroup$ – joseph Dec 10 '18 at 6:50
  • $\begingroup$ @joseph Because $\alpha =0$ in 2). $\endgroup$ – Kavi Rama Murthy Dec 10 '18 at 7:15

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