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!

  • $\begingroup$ @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

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 $n$.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.