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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 $a_n>0$ 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!

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  • $\begingroup$ @J.R. Well, $\;a_n=n\;$ doesn't fulfill that, yet $\;\sqrt[n]n\xrightarrow[n\to\infty]{}1\;$ ... $\endgroup$ – DonAntonio Nov 7 '16 at 20:38
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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 $e > 0$, and look at $b_n = (A+e)^n$. I claim that $b_n > a_n$ for all sufficiently large $n$. Proof: By definition of limit, we have that $|a_n^{1/n} - A| < \frac e2$ for all $n$ larger than some natural number $N$. But for any such $n$ we have $$ a_n^{1/n} < A+\frac{e}2 < A+e < b_n^{1/n}\\ a_n < b_n $$ This means that if $a_n^{1/n}$ converges, then there must be some number $B$ such that $a_n < B^n$, for all sufficiently large $n$, which again implies that there is such a $B$ that works for all $n$.

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  • $\begingroup$ Thanks for the answer and proof! Does the converse also hold, i.e. if there is $B$ such that $a_n<B^n$ for all $n$, then $a_n^{1/n}$ converges? $\endgroup$ – Szmagpie Nov 7 '16 at 20:56
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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.

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