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For every natural number $n$, let $$ x_n \colon= \frac{ \sqrt[n]{n!} }{ n }. $$ Then does the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ converge or diverge? And, how to find the limit?

My Attempt:

We find that, for any $n \in \mathbb{N}$, $$ \frac{ x_{n+1} }{ x_n } = \frac{ \frac{ \sqrt[n+1]{ (n+1)!} }{ n + 1 } }{ \frac{ \sqrt[n]{n!} }{ n } } = \frac{ \left( n! \right)^{ \frac{1}{n+1} - \frac{1}{n} } }{ 1 + \frac{1}{n} } \sqrt[n+1]{n+1} = . . . $$

What next? I was hoping to be able to apply the so-called "ratio" test for sequences, but there seems to be no such possibility available.

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    $\begingroup$ Try Stirling's approximation. $\endgroup$ – ProtectedSource Apr 20 '18 at 1:00
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    $\begingroup$ What is ratio test for sequences ? I keep seeing this on this site. Where did this idea come from ? $\endgroup$ – Rene Schipperus Apr 20 '18 at 1:03
  • $\begingroup$ @ReneSchipperus I asked myself the same question and Google sent me to this page: mathonline.wikidot.com/the-ratio-test-for-sequence-convergence $\endgroup$ – Malcolm Apr 20 '18 at 1:05
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    $\begingroup$ @Malcolm Well, thats some pretty low level math right there. $\endgroup$ – Rene Schipperus Apr 20 '18 at 1:07
  • $\begingroup$ If $x_{n+1}/x_n\to L < 1$ then $x_n \to 0$ for positive $x_n$ $\endgroup$ – Malcolm Apr 20 '18 at 1:07
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Consider $a_n = x_n^n = \dfrac{n!}{n^n}$. Then $$ \frac{a_{n+1}}{a_n} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{1+\frac1n}\right)^n \to \frac1e $$ Now, it is true that if $\lim \frac{a_{n+1}}{a_n}$ exists, then so does $\lim \sqrt[n]{a_n}$ and they are equal (see a proof here). Therefore, $$ \lim x_n = \lim \sqrt[n]{a_n} = \lim \frac{a_{n+1}}{a_n} = \frac1e $$

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  • $\begingroup$ Dont know why someone downvoted this, its nice. $\endgroup$ – Rene Schipperus Apr 20 '18 at 1:17
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Another way is to take logarithms. $$ \ln \frac{\sqrt[n]{n!}}{n}= \frac{1}{n}\sum_{k=2}^{n}{\ln k}-\ln n $$ Then using $$ \int_2^{n+1}{\ln x}\,\mathrm dx\ge \sum_{k=2}^{n}{\ln k} \ge \int_1^{n}{\ln x}\,\mathrm dx \\ \text{ and }\int{\ln x}\,\mathrm dx=x\ln x - x, $$ it's easy to see the logarithm goes to $-1$.

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Here is an easy trick, without sterling.

If $a_n\to L$ then $$\sqrt[n]{a_1a_2 \cdots a_n}\to L$$ apply this to $$a_n=\left(1+\frac{1}{n}\right)^n$$

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Using Stirling's approximation:

$$x_n\implies\frac{\sqrt[n]{n!}}{n}\sim\frac{(\tau n)^\frac1{2n}\left(\frac{n}{e}\right)^\frac{n}{n}}{n}\implies\frac{1}{e}(\tau n)^\frac1{2n}$$

$$\rightarrow\lim_{n\to\infty}\left(\frac{1}{e}(\tau n)^\frac1{2n}\right)=\boxed{\frac{1}{e}}$$

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    $\begingroup$ The limit of $n^{1/2n}$ is 1, not 0. $\endgroup$ – Mike Earnest Apr 20 '18 at 1:20
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Using Stirling's approximation: $$n!\sim\sqrt{2\pi n} \,\left(\frac{n}{e}\right)^n$$ You are looking for \begin{align} \lim_{n\to\infty} x_n&=\lim_{n\to\infty}\frac{\sqrt[n]{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}}{n}\\&=\lim_{n\to\infty} e^{-1} \left(2\pi n\right)^{1/(2n)} \\ &=\frac{1}{e}\lim_{n\to\infty} (2\pi)^{1/2n}\sqrt{n^{1/n}}\\ &=\frac{1}{e}\lim_{n\to\infty} (2\pi)^{1/2n} \cdot \sqrt{\lim_{n\to\infty} \sqrt[n]{n}}\\ &=\frac{1}{e} \end{align} So your sequences converges.

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  • $\begingroup$ Yes you are right. I edited the response. Thanks for telling me! $\endgroup$ – Zachary Apr 20 '18 at 1:21
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By Stirling approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

$$ x_n \colon= \frac{ \sqrt[n]{n!} }{ n }\sim \frac{ \sqrt[n]{\sqrt{2 \pi n}} }{ e }\to \frac1e$$

or as an alternative proceed by ratio-root criteria.

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