Radius of Convergence...very tricky question using very little information

Question

Even one of the maths gurus at my university struggled to get a proof out for this... so I'm almost completely lost!

This is the question: Let $a_{n}$ be a sequence of positive real numbers for which $s:=\lim_{n\rightarrow\infty}a_{n}^{\frac{1}{n}}$ exists. If $0<s<\infty$, show that the radius of convergence for:

$$\sum_{n=1}^{\infty}a_{n}x^{n}$$

is $\dfrac{1}{s}$. What happens in the case $s=0$ and $s=\infty$

Ok... so I could perform the ratio test, where:

$$r=\left|\dfrac{a_{n+1}x^{n+1}}{a_{n}x^{n}}\right|=\left|\dfrac{a_{n+1}x}{a_{n}}\right|$$

...which should somehow help us get to our solution of 1/s. Now, we have a power series, but this stuff is new enough that I'm completely at a loss as to how to use the limit to prove this result. Help would be greatly appreciated!

Answer 1

The radius of convergence is defined to be maximum value of $r$, such that $|x|<r\implies \sum{a_nx^n}$ converges then the series converges.

Use root test to arrive at the result.

Answer 2

Additionally, even if the root formula gives a limit, the ratio sequence does not have to converge. Just take $$ a_n=\begin{cases}2s^n&\text{ for odd }n\\s^n&\text{ for even }n\end{cases} $$

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Then consider $\limsup_{n\to\infty}\sqrt[n]{|a_nx^n|}<1$.