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Why is $\lim\limits_{n\to\infty}(n!)^{1/n}=\infty$

It is more or less clear that the sequence is increasing by ratio test $\frac{((n+1)!)^{1/(n+1)}}{(n!)^{1/n}}=\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n+1]{(n)!}}\cdot\frac{\sqrt[n+1]{(n)!}}{\sqrt[n]{(n)!}}=\sqrt[n+1]{n+1}\cdot(n!)^{\frac{1}{n(n+1)}}$

but left multiplicand tends to $1$, what about the right one ?

Is there convergence/divergence tests for sequences, I only found some for series ?

derivative
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  • The ratio and root tests work for sequences too. If the limit of ratio of absolute values is $<1$, the sequence converges to $0$; if it's $>1$, it diverges to $\infty$. –  Mar 10 '15 at 15:31
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    I think you can use the fact that $n!>\sqrt{n}^n$, and then the obvious fact that $\lim\limits_{n\to\infty}\sqrt{n}^n=\infty$. – barak manos Mar 10 '15 at 15:32
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    also use $n!^{\frac{1}{n}} = e^{\frac{\sum_k \log k}{n}} \geq e^{\frac{n \log n -n +1}{n}} = O(n) \to \infty$ – Alex Mar 10 '15 at 15:49
  • I mean, the obvious fact that $\lim\limits_{n\to\infty}(\sqrt{n}^n)^{1/n}=\infty$. – barak manos Mar 10 '15 at 16:05

5 Answers5

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It is obvious that in calculating n!, you are multiplying at least n/2 numbers which are greater than n/2.

$(n!)^{1/n} > ((n/2)^{n/2})^{1/n} = (n/2)^{1/2}$ which is clearly divergent.

gnasher729
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According to Stirling's approximation

$$n!\sim\sqrt{2n\pi}(\frac ne)^n$$ so $$\sqrt[n]{n!}\sim(2n\pi)^{\frac{1}{2n}}(\frac ne)$$ since $$\lim_{n\to \infty}(2n\pi)^{\frac{1}{2n}}=1$$

$viz.$ $$\lim_{n\to\infty}(2n\pi)^{\frac2n}(\frac ne)=\infty$$

Gabriel Romon
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Faye Tao
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Use that

$$a_n>0\;\;and\;\;a_n\xrightarrow[n\to\infty]{}L\implies \sqrt[n]{a_1\cdot\ldots\cdot a_n}\xrightarrow[n\to\infty]{}L\;:$$

$$\frac1n\xrightarrow[n\to\infty]{}0\implies\sqrt[n]{1\cdot\frac12\cdot\ldots\cdot\frac1n}=\frac1{\sqrt[n]{n!}}\xrightarrow[n\to\infty]{}0$$

and we're done

Timbuc
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Pick $n_0 \in \mathbb{N}$, then for $n \ge n_0$ we have $n! \ge n_0^n {n_0! \over n_0^{n_0}}$ and so $\sqrt[n]{n!} \ge n_0 \sqrt[n]{n_0! \over n_0^{n_0}}$.

Taking limits, we have $\liminf_n \sqrt[n]{n!} \ge n_0$. Since $n_0$ was arbitrary we have the desired result.

copper.hat
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Hint:

Show by induction that $$n!>\left(\frac{n}{e}\right)^n.$$

pointer
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