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Today I saw this limit and I was baffled by it.

$$e=\lim_{ n\to \infty} \sqrt[n]{\operatorname{lcm}(1,2,3,4,\ldots,n)}$$

Is there an elementary proof of the result?

curious
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  • First time I see this one, and it looks very interesting. I would doubt that there's any elementary proof of that. Perhaps you should start with asking what $\lim\limits_{n\to\infty}\sqrt[n]{n!}$ is equal to... BTW, it's not specified even on the definition of $e$ in Wikipedia (which lists quite a few alternative ways for calculating it). So if this is indeed true, then perhaps you want to add it there. – barak manos Feb 04 '15 at 17:02
  • Does this imply $\text{lcm}(1,2,\ldots,n)\to e^n$ as $n\to\infty$ ? – pshmath0 Feb 04 '15 at 17:05
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    Related http://math.stackexchange.com/questions/1111334/reason-for-lcm-of-all-numbers-from-1-n-equals-roughly-en – Caddyshack Feb 04 '15 at 17:08
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    @pbs: No. $e^n$ is a moving target, so nothing can tend to it. If you want to ask whether lcm$(1,2,...,n) - e^n$ tends to $0$, then the answer is still no. The ratio tends to $1$ (if the OP is to be believed), but that doesn't mean that the difference must tend to $0$, because both expressions grow without limit. – TonyK Feb 04 '15 at 17:09
  • @TonyK: According to the related question (linked one comment above yours), it actually does imply it (though it probably needs to be phrased with a different notation). – barak manos Feb 04 '15 at 17:10
  • @barak: I think you must be mistaken. For instance, $e^n$ is not, in general, especially close to an integer, whereas lcm$(...)$ is. Perhaps you didn't notice that $n$th roots are not being taken. – TonyK Feb 04 '15 at 17:12
  • @TonyK: I don't know, that's what the graph in that other question shows... – barak manos Feb 04 '15 at 17:13
  • @barak: Read my edited comment. – TonyK Feb 04 '15 at 17:14
  • @TonyK ok yes not a limit ("tends to") - my mistake. But maybe it is approximately $e^n$. – pshmath0 Feb 04 '15 at 17:21

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