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Suppose $a_n$ is an increasing unbounded positive sequence of real numbers. Let me specify $a_n$ a bit more for better understanding. It turns out that $a_n$ just being increasing unbounded positive real numbers does not suffice for a meaningful answer, as was pointed out in an answer.

Let $a_n=k_1k_2...k_n$ where $k_i$ are positive integers. Define $b_n=a_n^{1/n}$ and let $b:=\liminf_n b_n$. Is it true that $\inf_{n\geq1}\left[\dfrac{b_n}{b}\right]^n>0$?

[There was a lot of clutter in the post originally. Removing my attempt as it does not add anything at all for a basic start also]

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  • $\begingroup$ My apologies for not making explicit the nature of $a_n$. It's not really true that $a_n$ is ANY sequence. I am editing the post. $\endgroup$ – Landon Carter Oct 18 '16 at 13:43
  • $\begingroup$ The answer to the current version of the question is obviously "no", try $$a_n=\frac{3^n}n$$ $\endgroup$ – Did Oct 18 '16 at 15:16
  • $\begingroup$ Well, my $a_n$ has been defined to be the product of $n$ positive integers. $a_n=k_1k_2...k_n$. In this case, I am interested to know what happens. $\endgroup$ – Landon Carter Oct 18 '16 at 16:04
  • $\begingroup$ Yeah, then define $(k_n)$ recursively by $a_1=1$ and, for every $n\geqslant1$, $k_{n+1}=2$ if $a_n>\frac{3^n}{n+1}$ and $k_{n+1}=3$ otherwise, then $$\frac13\frac{3^n}n\leqslant a_n\leqslant\frac{3^n}n$$ for every $n$ hence the same conclusion applies. $\endgroup$ – Did Oct 18 '16 at 16:21
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Your question doesn't really make sense, since $b$ can be equal to $\infty$ (if $a_n=n^n$, for example) or $0$ (if $a_n=n$, for example). In both cases, $$\frac{b_n}{b}$$

is ill defined.

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  • $\begingroup$ My apologies for not making explicit the nature of $a_n$. It's not really true that $a_n$ is ANY sequence. I am editing the post $\endgroup$ – Landon Carter Oct 18 '16 at 13:46

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