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$$\lim_{n\rightarrow \infty}\sqrt[n]{n!}=\lim_{n\rightarrow \infty}\sqrt[n]{1}*\sqrt[n]{2}\cdots\cdot\sqrt[n]{n}=1\cdot1\cdot\ldots\cdot1=1$$ I already know that this is incorrect but I am wondering why. It probably has something to do with the fact that multiplication in $n!$ is done infinite number of times.

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  • $\begingroup$ Somewhat similar to this one. In that question that number of summands changes with $n$, in this question it is number of factors. $\endgroup$ – Martin Sleziak May 30 '13 at 12:12
  • $\begingroup$ Perhaps it is also appropriate to add at least one link to the correct proof. $\endgroup$ – Martin Sleziak May 30 '13 at 13:41
  • $\begingroup$ What's wrong is that $ \lim_{n\rightarrow \infty}{\sqrt[n]{1} \cdot \sqrt[n]{2} \cdots \sqrt[n]{n} = \lim_{n\rightarrow \infty}{1^n}}$ Which is an indeterminate form. $\endgroup$ – Bakuriu May 30 '13 at 14:49
  • $\begingroup$ $1^\infty$ is indeterminate. $\endgroup$ – A. Chu Jul 2 '13 at 13:07
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Start by figuring out a simpler example: $$1 = \lim_{n\to\infty} \frac n n = \lim_{n\to\infty} \frac {1+1+\ldots+1} n = \lim_{n\to\infty} \frac 1 n + \frac 1 n + \ldots + \frac 1 n = 0 + 0 + \ldots + 0 = 0$$

Indeed, you cannot exchange sum (or product) and limit if the amount of terms in the sum or product depend on the limiting variable.

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    $\begingroup$ $0\cdot\infty$ is indeterminate. $\endgroup$ – A. Chu Jul 2 '13 at 13:06
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Another way of explaining this is that for infinite $n$, each of the factors $\sqrt[n]{1}$, $\sqrt[n]{2}$, $\sqrt[n]{3}$, ... $\sqrt[n]{n}$ will be infinitely close to $1$, but this is not enough to conclude anything about the product because there are infinitely many factors in the product.

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