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Is it correct that $$\lim_{n->\infty}n!^{1/n}=\infty?$$ how to prove it using basic properties of sequence?

MJD
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4 Answers4

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Elementary proof (without Stirling)

In $n! = n\times (n-1) \times (n-2)\dots 2\times 1$, at least $\dfrac{n}{2}$ of the terms are $\geq \dfrac{n}{2}$, so $n! \geq \left(\frac{n}{2}\right)^\frac{n}{2}$.

This shows that $$n!^\frac{1}{n} \geq \sqrt{\frac{n}{2}}.$$

Siméon
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Another elementary proof: Split the sequence into pairs, 1 and $n$, 2 and $n-1$ and so on. The product of any pair is at least $n$, so the $n^{th}$ root of $n!$ is at least $\sqrt{n}$.

Empy2
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We have for large $n$ that $$n!\sim \left(\frac{n}{e}\right)^n \sqrt{2\pi n}$$ thus $$(n!)^\frac{1}{n} \sim \frac{n}{e}\left(2\pi n\right)^{\frac{2}{n}}.$$ Considering the two factors separately $$\lim_{n\rightarrow\infty}(2\pi n)^{\frac{1}{n}} = 1$$ and $$\lim_{n\rightarrow\infty} \frac{n}{e} = \infty$$ Thus the limit as $n\rightarrow\infty$ is $\infty$ not zero.

Dan
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  • Stirling's approximation is a pretty big hammer to use here, but on the other hand maybe that's a good thing. – MJD Oct 02 '13 at 15:57
  • It's the first thing that popped into my head; though I suppose technically it doesn't answer the question because it uses more than "basic properties of the sequence". – Dan Oct 02 '13 at 15:59
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    One nice thing about Stirling's approximation is that it squashes so many problems so very flat with such little effort. Everyone should have it in their toolbox, even if they don't plan to use it all the time. – MJD Oct 02 '13 at 16:03
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Dan's answer is correct, but you could also use the more general lemma:

Lemma: if $\;\{a_n\}\;$ is a positive sequence, then

$$\lim_{n\to\infty}a_n=A\implies \begin{cases}\lim_{n\to\infty}\sqrt[n]{a_1\cdot a_2\cdot\ldots\cdot a_n}=A\\{}\\\lim_{n\to\infty}\frac{a_1+a_2+\ldots+a_n}n=A\end{cases}$$

BTW, $\;A\;$ above can be finite or not.

DonAntonio
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