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I wanted to solve $\lim_{n \to \infty }(\frac{n!}{n^n})^{\frac{1}{n}}$ ( I saw many amazing solutions) but I was wondering where was my mistake in solving it

$$L=\lim_{n \to= \infty }(\frac{n!}{n^n})^{\frac{1}{n}}=\lim_{n \to \infty }((\frac{1}{n})(\frac{2}{n})(\frac{3}{n})(\frac{4}{n})(\frac{5}{n})......(\frac{n}{n}))^{\frac{1}{n}}$$ $$L=\lim_{n \to \infty }((1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})(1-\frac{4}{n})(1-\frac{5}{n})......(1-\frac{n-1}{n}))^{\frac{1}{n}}$$ $$L=\lim_{n \to \infty }((1-\frac{1}{n})^n(1-\frac{2}{n})^n(1-\frac{3}{n})^n(1-\frac{4}{n})^n(1-\frac{5}{n})^n......(1-\frac{n-1}{n})^n)^{\frac{1}{n^2}}$$ as limit of $(1-\frac{k}{n})^n=e^{-k}$ then $$L=\lim_{n \to \infty}(e^{-1} e^{-2}e^{-3}e^{-4}e^{-5}......e^{-n+1})^{\frac{1}{n^2}}$$ then $$L=\lim_{n \to \infty}(e^{-\frac{n(n-1)}{2n^2}})$$ then $L =e^{-0.5}$ which is wrong but what did I get wrong

it is probably making $\frac{1}{n}=e^{-n+1}$ but they have the same limit as n goes to infinity

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    One mistake is that you only have $n-1$ factors in the second line. The first factor would be $(1-\frac{0}{n})$, corresponding to the term $\frac{n}{n}$ in the first line. – MPW Jul 13 '23 at 20:36
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    "limit of $(1-\frac{k}{n})^{n}\to e^{-k}$" is true when $k$ is fixed. For example you claim that $(1-\frac{n-1}{n})^{n}\to e^{-(n-1)}$. But this is NOT a $1^{\infty}$ form but a $0^{\infty}$ form which has cannot be handled in that way. So in reality, it does not make any sense to consider the asymptotic regime to be $e^{-(n-1)}$. – Mr.Gandalf Sauron Jul 13 '23 at 20:37
  • @MPW I omitted $\frac{n}{n}$ –  Jul 13 '23 at 20:37
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    You cannot apply the limit transition theorem to a variable number of factors. There is a theorem that when there are limits to two terms, there is a limit to their product. This can be transferred to a fixed number of factors, but not to a variable number. – zkutch Jul 13 '23 at 20:45
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    Let'c consider counter example: obviously $n=\overbrace{\sqrt[n]{n} \cdots\sqrt[n]{n}}^{n\text{-times}}$, but knowing $\lim\limits_{n\to\infty}\sqrt[n]{n}=1$ we come to wrong result $$\infty=\lim\limits_{n\to\infty}n=\lim\limits_{n\to\infty}\overbrace{\sqrt[n]{n} \cdots\sqrt[n]{n}}^{n\text{-times}}=\overbrace{\lim\limits_{n\to\infty}\sqrt[n]{n} \cdots\lim\limits_{n\to\infty}\sqrt[n]{n}}^{n\text{-times}} = 1$$. – zkutch Jul 13 '23 at 21:11
  • but why this is not a violation to the rule $\lim(xy)=\lim(x) *\ lim(y)$ –  Jul 13 '23 at 21:35
  • @user 123. Read my first comment before counter example. – zkutch Jul 14 '23 at 00:08
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    Reminds me of a classic fake proof: $\frac{d}{dx}(x^2) = \frac{d}{dx} (x + \cdots \text{( x times )} \cdots + x) = 1 + \cdots \text{( x times )}\cdots + 1 = x$. – Charles Hudgins Jul 14 '23 at 01:49
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    @user123 Firstly $\lim(xy)=lim(x)*lim(y)$ holds if $\lim(x)$ and $\lim(y)$ both exists. Secondly you are just considering the product of two functions. In your case you are considering the product of a number $n$ of functions which is going to $\infty$. Just look at zkutch's example. If you are taking the product of a fixed number of functions, then you can apply the product rule if the individual limits exists. – Mr.Gandalf Sauron Jul 14 '23 at 07:30

2 Answers2

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"limit of $(1−\frac{k}{n})^{n}→e^{−k}$ is true when $k$ is fixed. For example you claim that $(1−\frac{n−1}{n})n→e^{−(n−1)}$. But this is NOT a $1^{\infty}$ form but a $0^{∞}$ form which has cannot be handled in that way. So in reality, it does not make any sense to consider the asymptotic regime to be $e^{−(n−1)}$.

What you can try is the following:

Assume the limit exists

Take log on both sides

Then $\displaystyle \log(L)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\log(\frac{k}{n})$ and now the RHS is a Riemann Sum which would give you

$\displaystyle\int_{0}^{1}\log(x)\,dx=-1$

So $\log(L)=-1\implies L=e^{-1}$

NOTE: I have not proved that the limit exists finitely. I am just showing that IF it existed, then it must be $e^{-1}$.

Alternatively look up Stirling's Approximation which gives you the asymptotics for $n!$. You can also use this directly to see that the limit is $e^{-1}$. That is,

$\dfrac{n!}{n^{n}}\sim \dfrac{n^{n}e^{-n}\sqrt{2\pi n}}{n^{n}}\sim e^{-n}\sqrt{2\pi n}$

Hence $\left(\dfrac{n!}{n^{n}}\right)^{\frac{1}{n}}\sim \left(e^{-n}\sqrt{2\pi n}\right)^{\frac{1}{n}}\to e^{-1} $

Mr.Gandalf Sauron
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  • can you explain to be why it can't be handled that way ? –  Jul 13 '23 at 21:36
  • and I am curious now how to prove the existence of a limit –  Jul 13 '23 at 21:37
  • @user123 Well, the first clue should be that you are ending up with the wrong answer. Secondly, $0^{\infty}$ is not an indeterminate form. So you cannot use the exponential asymptotics which holds for $1^{\infty}$ form. For example $(1-\frac{n-1}{n})^{n}=(\frac{1}{n})^{n}$ has much much faster decay than $e^{-n+1}$ as you are claiming them to be asymptotically equivalent. In fact $\lim_{n\to\infty}\frac{ (\frac{1}{n})^{n}}{e^{-n+1}}=0$ . So they CANNOT be asymptotic. – Mr.Gandalf Sauron Jul 14 '23 at 07:24
  • @user123 What do you mean by "curious"? I just gave you the proof in the "alternate" method I suggested using Stirling's approximation. In particular, one has to show that the sequence is "Cauchy" before computing the limit. That is what I meant by showing existence of the limit. But you can safely ignore such details at a high school level (which I presume you are) and directly proceed to computation of the limit. – Mr.Gandalf Sauron Jul 14 '23 at 07:33
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Taking log to the expression $q_n:=\Big(\frac{n!}{n^n}\Big)^{1/n}$ yields $$S_n=\frac{1}{n}\sum^{n-1}_{k=0}\log\big(1-\frac{k}{n}\big)$$

This is a Riemann sum for the improper integral $\int^1_0\log(1-x)\,dx=-1$

Since $-f(x)=-\log(1-x)$ is monotone increasing, the Riemann sum $S_n\xrightarrow{n\rightarrow\infty}-1$; hence $q_n\xrightarrow{n\rightarrow\infty}e^{-1}$.

This can also bee seem by looking at the ratio $r_n=\frac{a_{n+1}}{a_n}$ where $a_n=\frac{n!}{n^n}$. That is $$r_n=\frac{1}{\big(1+\frac{1}{n}\big)^n}\xrightarrow{n\rightarrow\infty}e^{-1}$$

It is a well known result that for a sequence $a_n>0$, $$\liminf_n\frac{a_{n+1}}{a_n}\leq\liminf_n\sqrt[n]{a_n}\leq\limsup_n\sqrt[n]{a_n}\leq \limsup_n\frac{a_{n+1}}{a_n} $$

Mittens
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