Suppose that there is an arbitrary $a_n$ and that $\sum_{n=1}^{\infty}a_n$ is a conditionally convergent series. Then prove if $\lim_{n \to \infty}\sqrt[n]{|a_n|}$ exists, then $\lim_{n \to \infty}\sqrt[n]{|a_n|}=1$.
Intuition
This seems to be a difficult question, since it is about an arbitrary series. What information should be utilized? I think the phrase conditionally convergent is key here.
Attempted Solution (sort of)
An example of a conditionally convergent series would be $\frac{(-1)^n}{n}$, and we know that $(\frac{1}{n})^{\frac{1}{n}}$ exists and converges to 1. From this example I see distinctly that if the absolute value of the $n^{th}$ square root of a conditionally convergent series exists, then alternating series are possible examples. But how would I exactly go about proving that every $a_n$ will have the property $\lim_{n \to \infty}\sqrt[n]{|a_n|}=1$?
Suppose that $\lim_{n\to\infty}\sqrt[n]{|a_n|}$ exists.
If it is greater than $1$, then $\sqrt[n]{|a_n|}>1$ if $n$ is large enogh, and therefore $|a_n|>1$ if $n$ is large enough. So, you don't have $\lim_{n\to\infty}a_n=0$, and therefore your series diverges.
And if it is smaller thatn $1$, the let $c\in\left(\sqrt[n]{|a_n|},1\right)$. Then $\sqrt[n]{|a_n|}<c$ is $n$ is large enough, and therefore $|a_n|<c^n$. So, your series converge absolutely.
So, $\lim_{n\to\infty}\sqrt[n]{|a_n|}=1$.