Suppose $r\lt1$. Prove that the series $\sum x_n$ is convergent. Let $x_n$ be a sequence in $\mathbb R$ and suppose $r = \lim_{n\rightarrow\infty} \root {n} \of {|x_n|}$ exists!
Suppose $r\lt1$. Prove that the series $\sum x_n$ is convergent.
I'm struggling to get started on this question, follow up questions include 'what if $r > 1$, $r = 0$', but after getting help with this one I should be able to pick it up.
My work so far:
$$\lim_{n\rightarrow\infty}|x_n| \le \lim_{n\rightarrow\infty}(|x_n|)^{\frac{1}{n}}\lt 1$$
Intuitively this to me means $\lim_{n\rightarrow\infty}|x_n| = 0$ since $r^n \rightarrow 0$ as $n\rightarrow \infty$. But this simply doesn't show a single thing when it comes to the convergence of the series of $x_n$.
Any guidance is appreciated!
 A: Your intuition is very close! Since $r<1$, we can pick a number $R$ with $r<R<1$. Now because $\sqrt{|a_n|}\to r$, can you show that there exists an $N$ such that $\sqrt{|a_n|}<R$ for all $n>N$? 
Let's see how to use this. 
We have 
$$
\sum_{n\geq1}|a_n|=\sum_{n=1}^N|a_n|+\sum_{n\geq N+1}|a_n|.
$$
Now, the first part on the right is some finite sum. For the second part, we know that $\sqrt{|a_n|}<R$, so $|a_n|<R^n$. Can you think of some test that we can apply this comparison towards to show that the right hand sum is also finite?
A: Note, clearly $0\le r$. By $r \lt 1$, we know there exists $R \in (r,1)$ such that $0\le r \lt R \lt 1$. Now choose $\epsilon = R - r$ then by hypothesis, there exists an $N$ such that $\forall n \ge N$ we have:
$$|\root n\of{|x_n|} - r| \le R - r$$
Hence $|\root n\of{|x_n|}| \le R$ for all $n \ge N$ (Triangle Inequality). Notice; $0 \le \root n\of{|x_n|} \le R$ and the cool thing is this is equivalent to:
$0 \le |x_n| \le R^n$ and $\sum R^n$ converges (geometric and $R \lt 1$), hence by comparison $\sum |x_n|$ converges!
