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This question already has an answer here:

Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$

So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$

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marked as duplicate by YuiTo Cheng, postmortes, Cesareo, Especially Lime, Paul Frost Jun 27 at 9:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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I would like to use the following lemma:

If $\lim_{n\to\infty}a_n=a$ and $a_n>0$ for all $n$, then we have $$ \lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}=a \tag{1} $$

Let $a_n=(1+\frac{1}{n})^n$, then $a_n>0$ for all $n$ and $\lim_{n\to\infty}a_n=e$. Applying ($*$) we have $$ \begin{align} e&=\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}\\ &=\lim_{n\to\infty}\sqrt[n]{\left(\frac{2}{1}\right)^1\left(\frac{3}{2}\right)^2\cdots\left(\frac{n+1}{n}\right)^n}\\ &=\lim_{n\to\infty}\sqrt[n]{\frac{(n+1)^n}{n!}}\\&= \lim_{n\to\infty}\frac{n+1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}+\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} \end{align}\tag{2} $$ where we use (1) in the last equality to show that $ \lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0. $

It follows from (2) that $$ \lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=\frac{1}{e}. $$

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  • $\begingroup$ Yes!! Thank you!! $\endgroup$ – Edgar Aroutiounian Sep 25 '12 at 0:39
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    $\begingroup$ I cannot follow what you are doing in your 3rd equal sign. $\endgroup$ – Martin Argerami Sep 25 '12 at 4:18
  • $\begingroup$ @MartinArgerami: Notice that $n!=1\cdot 2\cdots\cdot n$. $\endgroup$ – Jack Sep 25 '12 at 15:01
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    $\begingroup$ Yes, now I see it. I though it was a limit manipulation and not just algebra. Thanks. $\endgroup$ – Martin Argerami Sep 25 '12 at 15:03
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Have not found a way to rewrite your expression to get the desired result. However, here is a suggested approach.

Maybe rewrite the left-hand side as $$\sqrt[n]{\frac{n!}{n^n}}.$$

Take the logarithm. We get $$\frac{1}{n}\left(\log\left(\frac{1}{n}\right)+ \log\left(\frac{2}{n}\right)+\log\left(\frac{3}{n}\right)+\cdots+\log\left(\frac{n}{n}\right)\right).$$

Now think of the above sum as a Riemann sum for the not quite proper integral $$\int_0^1 \log x\,dx.$$

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    $\begingroup$ This is really neat! $\endgroup$ – Manny Reyes Sep 25 '12 at 0:01
  • $\begingroup$ This is very impressive. How did you come with it? Like what's the thought process behind the solution $\endgroup$ – Anurag Saha Jun 14 at 8:00
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If $a_n \geq 0$, then the following inequality holds:

$$ \liminf_{n\to\infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \frac{a_{n+1}}{a_n}. $$

Now let $ a_n = n! / n^n $. Then it follows that

$$ \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \frac{1}{\left(1+\frac{1}{n}\right)^n},$$

and hence

$$ \liminf_{n\to\infty} \frac{a_{n+1}}{a_n} = \limsup_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{1}{e}. $$

This proves that $\sqrt[n]{a_n} \to e^{-1}$ .

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  • $\begingroup$ For the inequality used in this answer see this post. (And also other posts listed there among linked questions.) $\endgroup$ – Martin Sleziak Jun 24 '14 at 7:56
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It's straightforward if you use Cesaro-Stolz theorem and then the celebre Lalescu's limit.

$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \lim_{n\to\infty} \sqrt[n+1]{(n+1)!}-\sqrt[n]{(n)!}=\frac{1}{e}.$$

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Here is a rather direct calculation of the limit using squeezing. It needs

  • $\ln k < \int_k^{k+1}\ln x \; dx < \ln (k+1)$
  • $ \int \ln x \; dx = x(\ln x - 1) \color{grey}{+C} $
  • $\lim_{n \rightarrow \infty}\frac{\ln (n+1)}{n} = 0$

Set $$ x_n = \ln \frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\sum_{k=1}^n \ln k - \ln n$$ So, to show is $\color{blue}{x_n \stackrel{n\rightarrow \infty}{\longrightarrow} -1}$. We have $$ \int_1^n \ln x \; dx < \sum_{k=1}^{n-1} \ln (k+1) =\sum_{k=1}^n \ln k < \int_1^{n+1}\ln x \; dx $$ We now squeeze: $$ \color{blue}{L_n} := \frac{1}{n}\int_1^n \ln x \; dx - \ln n \color{blue}{<x_n < } \frac{1}{n} \int_1^{n+1}\ln x \; dx - \ln n =: \color{blue}{R_n} $$

\begin{align*} \color{blue}{L_n} & = \frac{1}{n}\left( n(\ln n - 1) +1 \right) - \ln n \\ & = -1 +\frac{1}{n} \color{blue}{\stackrel{n \rightarrow \infty}{\longrightarrow} -1} \\ & \\ \color{blue}{R_n} & = \frac{1}{n}\left( (n+1)(\ln (n+1) - 1) +1 \right) - \ln n \\ & = \left( 1 + \frac{1}{n} \right) \ln (n+1) - \left( 1 + \frac{1}{n} \right) + \frac{1}{n} - \ln n \\ & = -1 + \ln \left( 1+\frac{1}{n} \right) + \frac{\ln (n+1)}{n} \color{blue}{\stackrel{n \rightarrow \infty}{\longrightarrow} -1} \end{align*}

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