Under what conditions does $\lim (a_n)^{1/n}$ converge? Under what conditions does $\lim (a_n)^{1/n}$ converge?  I was motivated by trying to solve limits and it would be quite useful to know if this converges and when $\lim(a_n)^{1/n}=\lim(a_n)$, or $\lim(a_n)^{1/n}=c$ for some constant $c$.

I have narrowed this down a little bit. This is what I have figured out so far: 


If $a_n$ is constant and $a_n&gt;0$ then this converges to $1$, if $a_n=0$, this converges to $0$,
If $a_n$ grows very large it clearly does not hold, for example if $a_n=e^{e^{e^n}}$ for all $n$ then the sequence diverges,
If $a_n=n$ for all $n$, or $a_n=cn$ for some constant $c$, then $\lim (a_n)^{1/n}=1$,
If $a_n$ is of the form $(b_n)^n$ then this will obviously converge iff $(b_n)$ does, and will share its limit; this way, for any $x\in\mathbb{R}_+$, I could construct sequences $(a_n)$ such that $\lim(a_n)^{1/n}=x$,
If $a_n$ is bounded and positive then this will converge to $1$.


I wonder whether there is a general rule for this?

(All limits in this question are taken as $n$ tends to $+\infty$.)

Thank you!
Assume that $a_n^{1/n}$ converges to some number $A$. As pointed out in the other answer, we must have $A \geq 0$, since otherwise the exponent $\frac1n$ doesn't make sense.

Now let $e &gt; 0$, and look at $b_n = (A+e)^n$. I claim that $b_n &gt; a_n$ for all sufficiently large $n$. Proof: By definition of limit, we have that $|a_n^{1/n} - A| &lt; \frac e2$ for all $n$ larger than some natural number $N$. But for any such $n$ we have
$$
a_n^{1/n} &lt; A+\frac{e}2 &lt; A+e &lt; b_n^{1/n}\\
a_n &lt; b_n
$$
This means that if $a_n^{1/n}$ converges, then there must be some number $B$ such that $a_n &lt; B^n$, for all sufficiently large $n$, which again implies that there is such a $B$ that works for all $n$.
Well, first $a_n$ has to be positive if $n$ is even, otherwise $a_n^{1/n}$ wouldn't make sense. Write down $a_n=e^{log(a_n)}$, then, since the exponential is continuous and has continuous inverse, it follows that $(a_n^{1/n})_{n\in\mathbb N}$ converges if and only if $\left(\frac{log(a_n)}{n}\right)_{n\in\mathbb N}$ does.