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Suppose $1<x_n\longrightarrow\infty$ is such that $\lim_{n\longrightarrow\infty}x_n^{\frac{1}{n}}=\rho>1$. What can we say about the growth of $x_n$, besides that $\log x_n \sim n\log \rho$? I was interested mainly in figuring out if there is anything to say, that speaks more directly about the slowest possible growth of $x_n$ without logarithms involved.

xyz
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    It's not necessarily the case that $\limsup x_n/\rho^n > 0$; for example, take $x_n = \rho^n / n$. – Greg Martin May 05 '22 at 21:10
  • Thank you you are right, gotta double check what I wrote, probably some careless mistakes hides somewhere. I remove that statement. – xyz May 05 '22 at 23:17
  • Fixed the mistake, what I actually get, also picking up from your hint, is that at the worst, if $x_n=o(\rho^n)$, rewriting $x_n=\rho^n\epsilon_n$ for $\epsilon_n\longrightarrow 0$, one has that $\log\varepsilon_n=o(n)$, which allows all sorts of polynomial dumping $\epsilon_n=1/n^k$ for any $k$... – xyz May 06 '22 at 01:53
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    The polynomial dumping is definitely correct; however the claims about $\epsilon_n$ are a little too broad (take $x_n = \rho^{n/2}$ or even $x_n=1$ for example). – Greg Martin May 06 '22 at 03:04
  • Sorry I forgot to mention I was only interested in the slowest possible regimes, I'll add it. I do not think $=^{n/2}$ would do as the limit of $x_n^{1/n}$ would be $\sqrt{\rho}$ and not $\rho$. Same with $=1$ which would have limit 1 but $\rho>1$. – xyz May 06 '22 at 11:19

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