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My attempt:$$\lim\limits_{n \to\infty}\sqrt[n]{n!}=\lim\limits_{n\to\infty}\sqrt[n]{1\cdot 2 \cdot 3 \cdot \ldots \cdot n}=\lim\limits_{n\to\infty}\sqrt[n]{1}\cdot \sqrt[n]{2}\cdot \sqrt[n]{3}\cdot \ldots \cdot \sqrt[n]{n}=1\cdot 1\cdot 1 \cdot \ldots \cdot 1=1$$ Why is this not correct?

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    $\begingroup$ So $\lim_{n\to\infty}1=\lim_{n\to\infty}(\frac1n+\frac1n+\cdots+\frac1n)=0+0+ \cdots+0=0$. $\endgroup$ – Lord Shark the Unknown Jun 16 at 15:01
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    $\begingroup$ Because there is no result that says that $$\lim_{n\to\infty}\prod_{k=1}^n a_{k,n}=\lim_{n\to\infty}\prod_{k=1}^n \lim_{m\to\infty} a_{k,m}$$ $\endgroup$ – Saucy O'Path Jun 16 at 15:04
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    $\begingroup$ The number of $1$s being multiplied together is $n$, and $n$ cannot appear outside of the limit (even if the product of $1$s doesn't actually depend on $n$). While each individual terms may approach $1$ in the limit, the product of $n$ of them, by virtue of accruing more and more factors, may diverge. $\endgroup$ – runway44 Jun 16 at 15:05
  • $\begingroup$ Lord's comment makes it all clear. Remember, you can not apply arithmetic of limits if the number of addends (or factors in a multiplication) depends on $\;n\;$ itself. $\endgroup$ – DonAntonio Jun 16 at 15:06
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    $\begingroup$ The arithmetic law of limits does not apply to infinitely many terms. $\endgroup$ – xbh Jun 16 at 15:10
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$n! = \prod_{k=1}^nk>\prod_{k=n/2}^n k>(n/2)^{n/2}$ so $(n!)^{1/n}>(n/2)^{1/2} \to \infty$.

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As the comments have pointed out, you are taking the limit of parts of the problem while ignoring the bigger picture, namely that you have to take the limit of the amount of parts as well.

Intuitively, we are saying the geometric mean of all positive integers is $\infty$. A simple proof may be done with ratio $\Rightarrow$ root:

$$\frac{(n+1)!}{n!}=n+1\to\infty\Rightarrow\sqrt[n]{n!}\to\infty$$

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  • $\begingroup$ Simply.Is your $\ge$ correct? Shouldn't it be $\le$? $\endgroup$ – Peter Szilas Jun 16 at 15:33
  • $\begingroup$ Fudge nuggets xd $\endgroup$ – Simply Beautiful Art Jun 16 at 15:35
  • $\begingroup$ Modified to a completely different approach. As an aside, it could have been fixed by showing the reciprocal approaches 0, though I find that a tad messier. $\endgroup$ – Simply Beautiful Art Jun 16 at 15:45
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It is incorrect to multiply the limits of each factor because the number of factors is not fixed, aech factor, which indeed tends to $1$ , is not equal to $1$, but slighly greater.

Hint: to prove the limit is $\infty$, you can try to determine the limit of the log, observing that $$\log(\sqrt[n]{n!})=\frac1n\bigl(\log 2+\dots+\log n\bigr),$$ which is an upper Riemann sum for the integral $\;\displaystyle\int_1^n \log x\,\mathrm dx$

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  • $\begingroup$ @DavidC.Ullrich: Sorry for the (freudian?) slip. It's fixed. Thank you for pointing it! $\endgroup$ – Bernard Jun 16 at 15:19
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By using the result that I proved here (https://math.stackexchange.com/a/3159844/629594), we have that $\lim\limits_{n\to \infty}\sqrt[n]{n!}=\lim\limits_{n\to \infty}\frac{(n+1)!}{n!}=\lim\limits_{n\to \infty}(n+1)=\infty$

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