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(Proof necessary)

$$\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n}$$

I don't have an answer yet, but I know it exists, and is less than $1$.

Edit. Winther's answer is the most correct I don't understand how he is jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). Don't presume, it's wrong, I need to go, and I'll keep looking at it when I get back

Any help is appreciated

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  • $\begingroup$ Is this your limit: $\lim_{n\rightarrow\infty} \frac{(n!)^{\frac{1}{n}}}{n}$? $\endgroup$ – Jared Sep 17 '14 at 20:15
  • $\begingroup$ Hint: If the limit exists, let's say its $L$, then $$\lim_{n \to \infty} \frac{n!}{n^n}=L^n$$ $\endgroup$ – Oria Gruber Sep 17 '14 at 20:18
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    $\begingroup$ Stirling's formula will help you $\endgroup$ – ThePortakal Sep 17 '14 at 20:18
  • $\begingroup$ See this question for something very similar. It can be done without Strilings formula. $\endgroup$ – Winther Sep 17 '14 at 20:32
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    $\begingroup$ @Oria: The value of a limit (if it exists) cannot depend on the limiting variable. So "$\displaystyle\lim_{n \to \infty}\frac{n!}{n^n} = L^n$" is definitely incorrect. $\endgroup$ – JimmyK4542 Sep 17 '14 at 20:51
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Put $$a_n = \frac{n!^{1/n}}{n}$$

then $$\log a_n = \frac{1}{n}\left(\log n! - n\log n\right)= \frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right)$$

where we have used $\log n! = \log 1 + \log 2 + \ldots + \log n$. The sum above is a Riemann sum for the integral $\int_0^1\log x dx$ so

$$\lim_{n\to\infty} \log a_n = \int_0^1\log x dx = [x\log x - x]_0^1 = -1$$

and it follows that $$\lim_{n\to\infty}a_n = \frac{1}{e}$$

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    $\begingroup$ Hey winther I don't understand how you're jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). $\endgroup$ – Calvin Hammond Sep 17 '14 at 21:28
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    $\begingroup$ @CalvinHammond $$\log n! - n \log n = \log 1 + \log 2 + \cdots + \log n - (\log n + \log n + \cdots + \log n),$$ where the number of terms in the parentheses is $n$. So $$\log n! - n \log n = \log \frac{1}{n} + \log \frac{2}{n} + \cdots + \log \frac{n}{n}$$ using the identity $\log x - \log y = \log \frac{x}{y}$. The rest follows. $\endgroup$ – heropup Sep 17 '14 at 21:42
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    $\begingroup$ Yes that is. Thanks for explaining it @heropup $\endgroup$ – Winther Sep 17 '14 at 21:46
  • $\begingroup$ Could you explain the remain sum more please. $\endgroup$ – Calvin Hammond Sep 17 '14 at 21:56
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    $\begingroup$ Notice that you have to be a little careful in applying this technique to improper integrals: math.stackexchange.com/questions/449103/… $\endgroup$ – user84413 Sep 18 '14 at 16:51
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Here is a simple elementary proof I found, but first of all, some lemmas:

  • This one could easily be proven by induction: $\displaystyle \prod_{k=1}^{n}\left(1+\frac{1}{k} \right) = n+1$

  • You can try to prove this inequality yourself since it's not difficult: $\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k} \right)^{k+1}\\$

  • This inequality is the one I'm going to use though because it gives a much tighter bound on our sequence and that's just more fun, though you could use the second inequality without change in proof. I could give you the proof if needed:$\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k} \right)^{k+1/2}\\$

Ok, so here is the proof:

  • We first write $n^n/n!$ in a better way: $\displaystyle \frac{n^n}{n!}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n \cdot \prod_{i=1}^{n-1}\prod_{k=1}^{i} \left(1+\frac{1}{k} \right)^{-1}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n\cdot \prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{-(n-i)}=\prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{i}$
  • Now, we use our inequalities to bound our sequence. First, an upper bound: $\displaystyle \prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{i} \leq \prod_{i=1}^{n-1}e^{1}=e^{n-1}$
  • Then, a lower bound: $\displaystyle \prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{i} \geq \prod_{i=1}^{n-1}\left (e^{1} \cdot \left(1+\frac{1}{i}\right)^{-1/2} \right )=\frac{e^{n-1}}{\sqrt[2]{n}}$
  • Now, since $\frac{n!^{1/n}}{n}=(\frac{n^n}{n!})^{-1/n}$, we get: $\displaystyle e^{\frac{1}{n}-1} \leq \frac{(n!)^{1/n}}{n} \leq e^{\frac{1}{n}-1} \cdot \sqrt[2n]{n}$
  • Finally, by the squeeze theorem, we get $\displaystyle e^{0-1} \leq \lim_{n \to \infty} \frac{(n!)^{1/n}}{n} \leq e^{0-1} \cdot 1$
  • Hence, $\displaystyle \lim_{n \to \infty} \frac{(n!)^{1/n}}{n}=e^{-1}$

I know there are simpler proofs, but this one is elementary and I feel like it gives you the direct intuition as to why it's true.

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Stirling's Approximation $$ n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$ Which means that $n!$ is asymptotically equivalent to $\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ as $n$ approaches $\infty$. If your familiar with asymptotic formulas, then you'd also know that this implies that $$ \lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n} =1 $$ Now, using the algebraic laws of limits, we have $$ \lim_{n\to\infty} n!=\lim_{n\to\infty} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$ $$ \lim_{n\to\infty} e^n=\lim_{n\to\infty} \sqrt{2\pi n}\left(\frac{n^n}{n!}\right) $$ $$ \lim_{n\to\infty} e=\lim_{n\to\infty} (2\pi n)^{\frac{1}{2n}}\left(\frac{n}{(n!)^{\frac{1}{n}}}\right) $$ So now $$ e=\left[\lim_{n\to\infty} e^{\ln(2\pi n)^{\frac{1}{2n}}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right] =\left[\lim_{n\to\infty} e^{\frac{\ln(2\pi n)}{2n}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right] $$ $$ =\left[\lim_{n\to\infty} e^{\frac{\frac{d}{dn}\ln(2\pi n)}{\frac{d}{dn}2n}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right] =\left[\lim_{n\to\infty} e^{\frac{1}{2n}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right] $$ $$ =e^0\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right]=1\cdot \lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}} $$ Thus $$ e=\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}} $$ And now we can easily see that $$ \lim_{n\to \infty} \frac{(n!)^{\frac{1}{n}}}{n}= \lim_{n\to \infty} \frac{1}{\left(\frac{n}{(n!)^{\frac{1}{n}}}\right)}=\frac{1}{e} $$ Let me know if you have any questions.

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Let $\displaystyle a_n=\frac{n!}{n^n}$.

Then $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\lim_{n\to\infty}\frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}$,

so $\;\;\;\;\displaystyle\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=\lim_{n\to\infty}\frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}$ $\hspace{.2 in}$ (since $\frac{a_{n+1}}{a_n}\rightarrow L\implies(a_n)^{\frac{1}{n}}\rightarrow L)$.

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    $\begingroup$ This is a great proof, but you would need to prove that the two forms of the limit are equivalent. $\endgroup$ – Fujoyaki Sep 17 '14 at 22:46
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    $\begingroup$ Thanks; you are right that I am using the result that if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L$, then $\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=L$. See math.stackexchange.com/questions/69386/… $\endgroup$ – user84413 Sep 17 '14 at 22:49
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Hint. Take the natural logarithm of $\dfrac{(n!)^{1/n}}{n}$ and obtain $$ \frac{\log 1+\log 2+\cdots+\log n}{n}-\log n=\frac{\log (1/n)+\log(2/n)+\cdots+\log(n/n)}{n} \\=\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right) \to \int_0^1 \log x\,dx=-1. $$

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Hint (Sterling): $$n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$$

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  • $\begingroup$ Yes that is my limit. I'm using this to help a friend but I don't know much about to how to solve this myself. So I'm asking for a full answer. $\endgroup$ – Calvin Hammond Sep 17 '14 at 20:28

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